Free math worksheets

- Test whether the first simplified statement can ever be true. - Remember that satisfying only one equation is not enough. - One impossible condition determines the fate of the whole system.

1. The statement \(0=5\) is false for every ordered pair. 2. A solution of a system must satisfy every equation simultaneously. 3. Since no ordered pair satisfies the first equation, the system has no solution regardless of the second equation.

The system has no solution.

- Compare the two rates of change. - Determine the geometric relationship implied by their product. - That relationship fixes how many times the lines can intersect.

1. The slopes are negative reciprocals, so the lines are perpendicular. 2. Perpendicular lines are not parallel and meet at exactly one point. 3. Therefore, the system has exactly one solution regardless of the intercepts.

The system has exactly one solution.

- Describe the graph of an equation that fixes only the x-coordinate. - Compare the two fixed values. - Vertical lines either coincide or never meet.

1. Each equation represents a vertical line. 2. If \(k=2\), the equations describe the same vertical line and have infinitely many solutions. 3. If \(k\ne2\), the vertical lines are distinct and parallel, so there is no solution.

For \(k=2\), the system has infinitely many solutions. For \(k\ne2\), it has no solution.

- Recall the geometric alternatives for two lines in a plane. - Separate coincident lines from distinct lines. - Identify which relationship is required for no intersection.

1. Two distinct nonparallel lines intersect at exactly one point. 2. Therefore, two nonparallel full lines cannot form a system with no solution.

No. Two nonparallel lines have exactly one common point, so their system has one solution.

- Compare the directions of the two lines. - Check whether the lines share any point. - A system’s solutions are the intersection points of its graphs.

1. Both graphs are vertical lines, so they have the same direction. 2. They have different x-values and therefore are distinct parallel lines. 3. The lines do not intersect, so the system has no solution.

The system has no solution.

- Simplify one equation before comparing the pair. - Check whether the variable coefficients become identical. - If the left sides match, compare what each equation requires on the right.

1. Divide equation (I) by \(3\): \(2x+3y=4\). 2. Equation (II) requires the same expression \(2x+3y\) to equal \(7\). 3. Because one expression cannot equal \(4\) and \(7\) at the same time, the system has no solution.

The solution set is empty; the system has no solution.

- Determine whether the two lines are parallel, identical, or intersecting. - A unique solution is represented by one intersection point. - Read both coordinates of the intersection from the grid.

1. Line \(a\) is vertical at \(x=-4\). Line \(b\) is not vertical, so the lines intersect exactly once. 2. The intersection shown on the graph is \((-4, 11)\). 3. Therefore, the system has exactly one solution.

The solution set is \(\{(-4, 11)\}\).

- Compare how the output changes over the same input interval. - Then compare the outputs at one shared input. - Equal rates can describe either the same line or distinct parallel lines.

1. For both lines, increasing \(x\) by \(2\) increases \(y\) by \(4\), so both slopes are \(2\). 2. At \(x=0\), Line A has \(y=3\) and Line B has \(y=8\), so the lines have different y-intercepts. 3. Distinct lines with equal slopes are parallel, so the system has no solution.

The system has no solution because the lines have the same slope but different y-intercepts.

- Rewrite the second equation in an equivalent simpler form. - Sharing one intercept does not determine whether two lines are identical. - Compare the complete equations, not just one point.

1. Divide the second equation by \(3\): \(y=-2x+5\). 2. Both equations represent the same line, not merely two lines sharing one point. 3. Every point on \(y=-2x+5\) satisfies both equations, so there are infinitely many solutions.

The flaw is treating equivalent equations as distinct lines. The system has infinitely many solutions.

- Read the rate of change from each point-slope equation. - Equal rates require one more comparison before classifying the lines. - Rewrite each equation in a form that makes its vertical position clear.

1. Both lines have slope \(\frac{3}{2}\). 2. Line \(a\) has y-intercept \(1\), since \(y=\frac{3}{2}x+1\). 3. Line \(b\) has y-intercept \(2\), since \(y=\frac{3}{2}x+2\). 4. The lines are distinct and parallel, so the system has no solution.

The system has no solution.

- Think about how many distinct lines can pass through two fixed points. - The two listed points are different, so they determine a line. - Decide whether the equations can represent separate lines under that condition.

1. Each linear equation represents a line. 2. There is exactly one line through two distinct points. 3. Since both lines pass through the same two distinct points, they must be the same line. 4. Therefore, every point on that line satisfies both equations.

The system has infinitely many solutions.

- One intersection point represents one ordered-pair solution. - Distinct parallel lines never meet. - When two equations are represented by one shared line, every point on that line satisfies both equations.

1. In graph a), the lines intersect once at \((1,2)\), so the system has exactly one solution. 2. In graph b), the lines are distinct and parallel, so the system has no solution. 3. In graph c), both equations are represented by the same solid line, so the system has infinitely many solutions.

a) Exactly one solution: \(\{(1,2)\}\). b) No solution. c) Infinitely many solutions.

- Compare whether one pair of variable coefficients is a common multiple of the other. - Proportional coefficient pairs are the special case to watch for. - Distinct slopes determine how many times two lines can meet.

1. The ratio of the x-coefficients is \(\frac{7}{3}\), while the ratio of the y-coefficients is \(\frac{-2}{4}=-\frac{1}{2}\). 2. Because the coefficient pairs are not proportional, the equations represent lines with different slopes. 3. Two nonparallel lines intersect exactly once, so the system has one solution.

The system has exactly one solution.

- Compare the output changes across consecutive rows. - Look for an input at which the two outputs match. - Different rates prevent the lines from sharing more than one point.

1. Line A changes by \(3\) in \(y\) for each increase of \(1\) in \(x\), while Line B changes by \(-1\). 2. The slopes are different, so the lines have exactly one intersection. 3. The table shows that both lines contain \((0, 5)\), so that is the unique solution.

The system has exactly one solution, \((0, 5)\).

- Remove the denominators using a common multiple. - Compare the complete equations after rewriting. - Identical equations impose only one line of restrictions.

1. Multiply equation (I) by \(12\): \(3x+2y=12\). 2. This is exactly equation (II). 3. The equations represent the same line, so the system has infinitely many solutions.

The system has infinitely many solutions.

- Analyze each candidate against the first equation on its own. - If the variable coefficients are proportional, separately compare the constants. - If the coefficient pairs are not proportional, what does that imply about the slopes?

1. For a), every coefficient and the constant are twice those in \(3x-y=6\), so the equations represent the same line. The system has infinitely many solutions. 2. For b), the variable coefficients are doubled but the constant is not, so the equations represent distinct parallel lines. The system has no solution. 3. For c), the coefficient pair \((6,1)\) is not proportional to \((3,-1)\), so the lines have different slopes. The system has exactly one solution.

a) Infinitely many solutions; the second equation is exactly twice the first. b) No solution; the variable coefficients are proportional but the constants are not. c) Exactly one solution; the coefficient pairs are not proportional.

- Determine whether any equation repeats information already present. - Compare the equations that remain after removing redundancy. - A common solution must satisfy all three equations at once.

1. Dividing equation (II) by \(3\) gives \(x+2y=4\), so equation (II) is redundant with equation (I). 2. Equation (III) requires the same expression \(x+2y\) to equal \(7\) instead of \(4\). 3. No ordered pair can satisfy both requirements, so the three-equation system has no solution.

The system has no solution. Equation (II) repeats equation (I) and does not remove the contradiction with equation (III).

- Picture two lines with the same rate of change. - Decide what happens when their vertical positions differ. - Then consider the special case when their vertical positions also match.

1. Equal slopes mean the lines cannot cross at exactly one point. 2. If the lines have different intercepts, they are distinct and parallel, so there is no solution. 3. If they have the same intercept, they are the same line, so there are infinitely many solutions.

No solution and infinitely many solutions are possible; exactly one solution is not. Comparing an intercept determines which case occurs.

- Scale one equation exactly and compare every term. - “Close” constants do not create a common point when the slopes match. - Decide whether one expression can equal two different exact numbers.

1. Doubling equation (I) gives \(6x-14y=2\). 2. Equation (II) has the same variable coefficients but requires the left side to equal \(2.1\). 3. The equations represent distinct parallel lines, so they never intersect. The difference of \(0.1\) changes the solution type from coincident to parallel.

The system has no solution. Exact proportionality of all coefficients is required for coincident lines.

- Standardize the term order before comparing coefficients. - A negative scale factor changes every term, including the constant. - Equivalent equations represent one shared line.

1. Multiply equation (I) by \(-2\): \(-2y+4x=-8\). 2. Reordering gives \(4x-2y=-8\), exactly equation (II). 3. The equations represent the same line, so the system has infinitely many solutions.

The system has infinitely many solutions.

- One equation already fixes one coordinate completely. - Check whether the remaining equation can always determine the other coordinate. - Look for any parameter value that would make the remaining coefficient vanish.

1. Equation (II) fixes the y-coordinate. 2. Substitute \(y=2\) into equation (I): \(x+2k=3\), so \(x=3-2k\). 3. This calculation works for every real \(k\), so the system always has exactly one solution.

For every real \(k\), the solution set is \(\{(3-2k, 2)\}\).

- Start with the equation that fixes a coordinate directly. - A candidate from two equations must still satisfy every remaining equation. - Verify the same ordered pair across the full system.

1. From equation (III), \(x=2\). 2. Substitute into equation (I): \(2+y=5\), so \(y=3\). 3. The pair \((2, 3)\) also satisfies equation (II), since \(2\cdot2-3=1\). 4. All three equations share exactly this one point.

The solution set is \(\{(2, 3)\}\).

- Look for a value that makes the shared factor vanish. - Compare the coefficients multiplying that factor. - A common point plus different slopes determines the solution type.

1. Both equations equal \(0\) when \(x=4\), so both lines pass through \((4, 0)\). 2. Their slopes are \(2\) and \(-5\), which are different. 3. Lines with different slopes share only one point, so \((4, 0)\) is the unique solution.

The solution set is \(\{(4, 0)\}\).

- A point must lie on both line graphs to solve the system. - Locate the single point where the two lines cross. - Use the intersection pattern to classify the number of solutions.

1. A solution of the system must lie on both lines. 2. Point \(A\) is at the intersection \((2, 3)\), so it satisfies both equations represented by the lines. 3. The lines have different slopes and intersect once, so the system has exactly one solution.

Point \(A\) is the solution, and the solution set is \(\{(2, 3)\}\).

- Start with what the known common point rules out. - Consider separately whether the two lines are distinct or identical. - A single shared point does not prove the lines are different.

1. Because \((3, -1)\) satisfies both equations, the system cannot have no solution. 2. If the equations represent distinct lines, they intersect only at \((3, -1)\), giving exactly one solution. 3. If the equations represent the same line, they have infinitely many solutions. 4. Therefore, one solution and infinitely many solutions are possible; no solution is impossible.

Exactly one solution or infinitely many solutions are possible. No solution is impossible.

- Rewrite the equation to identify its rate of change. - Find where the equation crosses the x-axis. - A line is fixed by its slope and one point.

1. Line B can be written as \(y=3x+6\), so its slope is \(3\). 2. Setting \(y=0\) gives \(3x=-6\), so its x-intercept is \(-2\). 3. The lines have the same slope and the same x-intercept, so they are the same line.

The system has infinitely many solutions.

- Check whether one entire equation is a nonzero multiple of the other. - Visual appearance is less important than the solution set each equation defines. - A whole-equation scaling preserves every solution.

1. Multiply the first equation by \(-3\): \(-6x+15y=-21\). 2. This is exactly the second equation. 3. The equations look different but represent the same line, so the system has infinitely many solutions and the claim is false.

The claim is false; the two equations are equivalent and the system has infinitely many solutions.

- Find what coordinate the first two equations force. - Compare that requirement with the third equation. - A point satisfying two equations may fail when another condition is added.

1. Add equations (I) and (II): \(3x=9\), so their intersection has \(x=3\). 2. Equation (III) requires \(x=\frac{8}{3}\). 3. No ordered pair can have both x-coordinates, so the three-equation system has no solution.

The common solution set is empty.

- Decide whether each final statement is always true, never true, or determines a value. - A valid elimination result preserves the original solution set. - Distinguish an identity from a contradiction.

1. The identity \(0=0\) shows that the equations are dependent, so the system has infinitely many solutions. 2. The contradiction \(0=9\) cannot be satisfied, so the system has no solution. 3. The equation \(5x=15\) determines \(x=3\). The stated original equation then determines one value of \(y\), so the system has exactly one solution.

a) Infinitely many solutions b) No solution c) Exactly one solution

- A shared point is only one feature of a line. - Compare the rates of change before deciding the lines coincide. - Distinct slopes determine the maximum number of shared points.

1. The slopes are \(1\) and \(-2\), so the equations represent distinct lines. 2. Both lines contain \((2,3)\), making it their intersection point. 3. Distinct lines with different slopes share exactly one point.

The lines are not the same. The solution set is \(\{(2,3)\}\).

- An equation operation must affect both sides equally. - Compare the correct scaled constant with the stated one. - Proportional left sides can still describe different parallel lines.

1. Multiplying all of equation (I) by \(3\) would give \(3x+6y=12\), not equation (II). 2. The actual equations have proportional variable coefficients but inconsistent constants. 3. They represent distinct parallel lines, so the system has no solution.

The constant also needed to be multiplied by \(3\). The actual system has no solution.

- Translate the constant output difference into a vertical relationship. - Compare how both outputs change when the input changes. - A common solution would require the output difference to equal zero.

1. The relation \(g(x)-f(x)=4\) means the graph of \(g\) is always \(4\) units above the graph of \(f\). 2. Because the vertical difference is constant, the lines have the same slope. 3. The outputs can never be equal for the same input, so the lines do not intersect.

The lines have the same slope and a constant vertical separation of \(4\), so the system has no solution.

- Move variable terms and constants to reveal each equation’s simplest form. - Compare the simplified left sides first. - Matching left sides with different constants create a contradiction.

1. Equation (I) simplifies to \(x+y=-4\). 2. Equation (II) simplifies to \(x+y=2\). 3. The same expression cannot equal \(-4\) and \(2\) simultaneously, so the system has no solution.

The system has no solution.

- Distinguish having the same slope from being distinct parallel lines. - Ask whether two separate lines with the same slope can share a point. - Use the shared point to resolve the ambiguity.

1. Distinct lines with the same slope cannot share any point. 2. Since both lines contain \((-4, 5)\), they cannot be distinct. 3. Therefore, the descriptions refer to the same line, which gives infinitely many solutions.

The system has infinitely many solutions.

- Undo the operation used to create the second equation. - An operation performed equally on both sides preserves an equation’s solutions. - Decide how many distinct line conditions remain.

1. Subtracting \(4\) from both sides of the new equation recovers \(y=mx+b\). 2. The transformation preserves every solution of the original equation. 3. The two equations describe the same line for all real \(m\) and \(b\), so the system has infinitely many solutions.

The system has infinitely many solutions for all real \(m\) and \(b\).

- Translate the verbal coordinate relationship into an equation. - Rearrange one form so the terms appear in the same order. - Decide whether the descriptions impose one or two distinct conditions.

1. The verbal rule for Line A is \(y=x+3\). 2. Rearranging gives \(x-y=-3\), exactly the equation of Line B. 3. The lines coincide, so the system has infinitely many solutions.

The system has infinitely many solutions.

- Use the given point to determine the missing constant. - After completing the equation, compare all coefficients with the first line. - A common scale factor across every term indicates one line.

1. Substitute \((0, 3)\) into Line B: \(c=6\cdot0+8\cdot3=24\). 2. Line B is \(6x+8y=24\), which is twice Line A. 3. The lines coincide, so the system has infinitely many solutions.

\(c=24\), and the system has infinitely many solutions.

- Divide the first equation by \(4\). - Compare the left sides of the two equations. - When are two lines identical, and when are they parallel? - Could lines with the same slope intersect exactly once?

1. Divide equation (I) by \(4\): \(3x - y = 2\). 2. When \(k = 2\), equation (II) is identical to equation (I), so the system has infinitely many solutions. 3. When \(k \ne 2\), the equations have identical left sides but different constants. They represent distinct parallel lines, so the system has no solution. 4. For every value of \(k\), the lines have the same slope. They are therefore either identical or parallel and can never intersect at exactly one point.

a) \(k = 2\) b) \(k \ne 2\) c) The lines always have the same slope, so they are either identical or distinct parallel lines.

- Compare the rates of change determined by each pair of points. - Equal rates alone are not enough; compare where each line crosses an axis. - Decide whether the lines are distinct or identical.

1. The slope of \(r\) is \(\frac{7-1}{4-(-2)}=1\). Using \((-2, 1)\), its equation is \(y=x+3\). 2. The slope of \(s\) is \(\frac{8-3}{5-0}=1\). Since it passes through \((0, 3)\), its equation is also \(y=x+3\). 3. The two equations represent the same line, so every point on that line satisfies both equations.

The system has infinitely many solutions because both equations represent \(y=x+3\).

- Put both rules in the same form before comparing them. - Separate the algebraic line from the restrictions imposed by the situation. - Decide which input values are meaningful for counted hours.

1. Divide Rule B by \(3\): \(C=12h+25\), which is exactly Rule A. 2. Algebraically, the two equations represent the same line and have infinitely many real ordered-pair solutions. 3. In context, \(h\) must be a nonnegative whole number, and each such value gives \(C=12h+25\).

a) The algebraic system has infinitely many solutions. b) The contextually valid solutions are \(\left\{(h, 12h+25)\mid h\in\{0, 1, 2, \ldots\}\right\}\).

- Compare the coefficients that already match. - Ask when the remaining coefficient also matches. - In the nonmatching case, combine the equations to see which coordinate is forced.

1. The equations are identical when \(k=3\), so that case has infinitely many solutions. 2. For \(k\ne3\), subtract equation (II) from equation (I): \((k-3)x=0\). Since \(k-3\ne0\), \(x=0\). 3. Substituting \(x=0\) gives \(2y=4\), so \(y=2\). Thus, every other value of \(k\) gives the unique solution \((0, 2)\).

a) \(k=3\), giving infinitely many solutions. b) For \(k\ne3\), the solution set is \(\{(0, 2)\}\).

- Express the shifted line using the original rule. - Compare the rate of change before comparing the vertical positions. - A nonzero horizontal shift of a nonhorizontal line can create a parallel line.

1. A horizontal shift right by \(6\) replaces \(x\) with \(x-6\): \(y=\frac{1}{2}(x-6)-4\). 2. This simplifies to \(y=\frac{1}{2}x-7\). 3. The lines have the same slope and different y-intercepts, so they are parallel and distinct.

The system has no solution.

- Consider separately what happens when the slope is zero. - For the nonzero case, compare the two slopes and a point both equations share. - A reflection can leave a line unchanged only in a special case.

1. When \(m=0\), both equations are \(y=b\), so they represent the same horizontal line and have infinitely many solutions. 2. When \(m\ne0\), the slopes \(m\) and \(-m\) are different. 3. Both lines pass through \((0, b)\), so for \(m\ne0\) they intersect exactly once at \((0, b)\).

For \(m=0\), the system has infinitely many solutions. For \(m\ne0\), it has exactly one solution, \((0, b)\).

- Compare what changes when the variables are exchanged. - Combining the equations can reveal a relationship between the coordinates. - Use that relationship to locate the common point.

1. Subtract equation (II) from equation (I): \(-3x+3y=0\), so \(x=y\). 2. Substitute \(y=x\) into either equation: \(7x=9\), so \(x=\frac{9}{7}\) and \(y=\frac{9}{7}\). 3. The system therefore has exactly one solution.

The solution set is \(\left\{\left(\frac{9}{7}, \frac{9}{7}\right)\right\}\).

- Convert each pair of intercepts into two points. - Compare the rates of change using those points. - Equal rates require checking whether the lines occupy the same position.

1. Line A passes through \((2, 0)\) and \((0, -4)\), so its slope is \(\frac{0-(-4)}{2-0}=2\). 2. Line B passes through \((5, 0)\) and \((0, -10)\), so its slope is \(\frac{0-(-10)}{5-0}=2\). 3. The y-intercepts are different, so the lines are distinct and parallel.

The system has no solution.

- For one case, preserve every ratio from the first equation. - For another case, preserve only the variable-coefficient ratio. - For the unique case, choose a different line through the required point.

1. For infinitely many solutions, a valid choice is \(4x+6y=12\), which is twice the first equation. 2. For no solution, a valid choice is \(4x+6y=15\), which has proportional variable coefficients but an inconsistent constant. 3. For one solution at \((0, 2)\), a valid choice is \(x=0\). It passes through \((0, 2)\) and is not parallel to the first line.

a) One answer is \(4x+6y=12\). b) One answer is \(4x+6y=15\). c) One answer is \(x=0\).

- Compare the per-page rates before finding the crossing point. - At a common point, both cost expressions have the same value. - Interpret both coordinates using the quantities named in the problem.

1. The rates \(0.08\) and \(0.05\) are different, so the lines intersect exactly once. 2. Set the costs equal: \(0.08p+20=0.05p+35\). 3. Then \(0.03p=15\), so \(p=500\). Substitution gives \(C=60\). 4. Both services cost the same for a \(500\)-page job, and that common cost is \(\$60\).

a) The system has exactly one solution. b) The solution set is \(\{(500, 60)\}\): both services charge \(\$60\) for \(500\) pages.

- Separate the case of a horizontal second line from a slanted one. - Two horizontal lines depend on whether their heights match. - A nonhorizontal line crosses the x-axis exactly once.

1. If \(m=0\) and \(b=0\), both equations are \(y=0\), so there are infinitely many solutions. 2. If \(m=0\) and \(b\ne0\), the lines are distinct horizontal lines, so there is no solution. 3. If \(m\ne0\), the second line is nonhorizontal and crosses the x-axis once. Solving \(0=mx+b\) gives the unique solution \(\left(-\frac{b}{m}, 0\right)\).

a) Infinitely many solutions b) No solution c) Exactly one solution, \(\left(-\frac{b}{m}, 0\right)\)

- Check whether the parameter changes the directions of the lines or only their positions. - The fixed coefficient pairs determine whether parallelism is possible. - Symmetric combinations of the equations isolate the coordinates.

1. The coefficient pairs \((1, 1)\) and \((1, -1)\) are not proportional, so the lines are never parallel. 2. Add equations (I) and (II): \(2x=k+2\), so \(x=\frac{k}{2}+1\). 3. Subtract equation (II) from equation (I): \(2y=k-2\), so \(y=\frac{k}{2}-1\). 4. Thus, every real \(k\) gives exactly one solution.

For every real \(k\), the solution set is \(\left\{\left(\frac{k}{2}+1, \frac{k}{2}-1\right)\right\}\).

- Find the rate of change shown by the table. - Use one table point to determine the intercept. - Compare the resulting rule with the given equation.

1. The outputs decrease by \(4\) when \(x\) increases by \(2\), so Line B has slope \(-2\). 2. Using \((1, 4)\), its y-intercept is \(6\), so Line B is \(y=-2x+6\). 3. The two lines are the same, so the system has infinitely many solutions.

The system has infinitely many solutions.

- Convert each pair of intercepts into an equation. - Notice the symmetry created by swapping the intercepts. - A simple combination can show how the coordinates are related.

1. Line A has equation \(\frac{x}{4}+\frac{y}{2}=1\), or \(x+2y=4\). 2. Line B has equation \(\frac{x}{2}+\frac{y}{4}=1\), or \(2x+y=4\). 3. The coefficient pairs are not proportional, so the system has one solution. 4. Subtracting the equations gives \(x-y=0\). Thus, \(x=y\), and \(3x=4\), so \(x=y=\frac{4}{3}\).

The solution set is \(\left\{\left(\frac{4}{3}, \frac{4}{3}\right)\right\}\).

- Treat the zero and nonzero parameter cases separately. - Division by the parameter is valid in only one case. - An identity does not remove any points from the first line.

1. If \(r\ne0\), dividing equation (II) by \(r\) gives \(2x-y=6\), so the equations are identical. 2. If \(r=0\), equation (II) becomes \(0=0\), which adds no restriction to equation (I). 3. In both cases, every point on \(2x-y=6\) satisfies the system.

For every real \(r\), the system has infinitely many solutions.

- Track both a point on the line and the line’s direction under the rotation. - A line is determined by one point and a direction. - Decide whether the image can be a distinct parallel line when the center lies on the original.

1. A \(180^\circ\) rotation sends every direction along a line to the opposite direction along the same line. 2. Because the center \(P\) lies on \(L\), the rotated image passes through \(P\) and has the same line direction. 3. Thus \(M\) is the same line as \(L\), so the system has infinitely many solutions.

The system has infinitely many solutions.

- Compare the complete left sides before considering the constants. - One fixed expression cannot take two different values at the same ordered pair. - The condition on \(a\) and \(b\) ensures the equation represents a line.

1. If \(c=d\), the equations are identical and describe one line, so there are infinitely many solutions. 2. If \(c\ne d\), the same expression \(ax+by\) is required to equal two different constants. 3. That is impossible, so the system has no solution in the second case.

If \(c=d\), there are infinitely many solutions. If \(c\ne d\), there is no solution.

- When do two equations represent the same line, and when do they represent parallel lines? - Compare corresponding coefficients and constants. - Equal or proportional left sides with nonproportional constants indicate no solution. - To create an equation through a point, substitute its coordinates into a simple linear form.

1. Divide equation (I) by \(2\): \(2x - 3y = 6\). Therefore, when \(k = 6\), equations (I) and (II) represent the same line and the system has infinitely many solutions. 2. When \(k = 10\), equation (II) is \(2x - 3y = 10\). Multiplying by \(2\) gives \(4x - 6y = 20\). This line has the same left side as equation (I) but a different constant, so the lines are parallel and the system has no solution. 3. A valid equation through \((3, 0)\) that is not a multiple of equation (I) is \(x + y = 3\). Together with equation (I), it has the unique solution \((3, 0)\).

a) \(k = 6\) b) The equations represent distinct parallel lines when \(k = 10\), so there is no solution. c) One possible equation is \(x + y = 3\).

- Solve the second equation for one variable and substitute. - Identify the expression that would be used as a divisor. - What happens when that expression equals zero?

1. Solve equation (II) for \(y\): \(y = 2x - 3\). 2. Substitute into equation (I): \(ax + 5(2x - 3) = 15\). 3. Simplify: \((a + 10)x = 30\). 4. If \(a \ne -10\), divide to get \(x = \frac{30}{a + 10}\). 5. Then \(y = 2x - 3 = \frac{30 - 3a}{a + 10}\). 6. If \(a = -10\), the equation becomes \(0 = 30\), so the system has no solution.

For \(a \ne -10\), \((x, y) = \left(\frac{30}{a + 10}, \frac{30 - 3a}{a + 10}\right)\). For \(a = -10\), the system has no solution.

- Scale the first equation so that the \(y\)-coefficients match. - What happens when the \(x\)-coefficients also match? - If they do not match, what must \(x\) equal?

1. Multiply equation (I) by \(3\): \(3x + 6y = 18\). 2. Subtract this equation from equation (II): \((m - 3)x = 0\). 3. If \(m = 3\), the original equations are equivalent, so the system has infinitely many solutions. 4. If \(m \ne 3\), then \(x = 0\). Substitute into equation (I): \(2y = 6\), so \(y = 3\). Thus, the unique solution is \((0, 3)\).

If \(m = 3\), the system has infinitely many solutions. If \(m \ne 3\), the system has the unique solution \((0, 3)\).

- Use substitution and note when a variable disappears. - When does the resulting equation become a contradiction? - What would be required for an identity?

1. Solve equation (II) for \(x\): \(x = 3y + 5\). 2. Substitute into equation (I): \(2(3y + 5) + ay = 8\), so \((a + 6)y = -2\). 3. If \(a = -6\), the equation becomes \(0 = -2\), a contradiction. Therefore, the system has no solution. 4. Infinitely many solutions would require an identity such as \(0 = 0\). Because the right side remains \(-2\), no value of \(a\) produces infinitely many solutions.

The system has no solution when \(a = -6\). There is no value of \(a\) for which it has infinitely many solutions.

- Scale one equation so the \(y\)-coefficients match. - When do matching variable coefficients create a contradiction? - When do they create equivalent equations? - What slope relationship guarantees one intersection?

1. Multiply equation (I) by \(-2\): \(-4x + 6y = -12\). 2. If \(a = -4\), the variable coefficients match. 3. For no solution, the constants must differ: \(a = -4\) and \(b \ne -12\). 4. For infinitely many solutions, the equations must be identical: \(a = -4\) and \(b = -12\). 5. For exactly one solution, the slopes must differ: \(a \ne -4\), with any value of \(b\).

1. \(a = -4\), \(b \ne -12\) 2. \(a = -4\), \(b = -12\) 3. \(a \ne -4\), with any \(b\)

- Eliminate one variable while keeping \(k\) in the coefficients. - For which values of \(k\) does the resulting variable coefficient become zero? - Substitute each resulting value of \(k\) into the elimination equation. - Does elimination produce an identity or a contradiction?

1. Multiply equation (I) by \(k + 1\): \(2(k + 1)x + (k^2 - 1)y = 6(k + 1)\). 2. Multiply equation (II) by \(2\): \(2(k + 1)x + 8y = 24\). 3. Subtract the second transformed equation from the first: \((k^2 - 9)y = 6k - 18\). 4. The system can fail to have exactly one solution only when the coefficient of \(y\) is zero: \(k^2 - 9 = 0\). Thus, \(k = 3\) or \(k = -3\). 5. If \(k = 3\), the elimination equation is \(0 = 0\). The original equations are equivalent, so there are infinitely many solutions. 6. If \(k = -3\), the elimination equation is \(0 = -36\), a contradiction, so there is no solution.

For \(k = 3\), the system has infinitely many solutions. For \(k = -3\), the system has no solution.

- Eliminate one variable while keeping \(a\) in the coefficients. - For which values of \(a\) does the resulting variable coefficient become zero? - Substitute each candidate value into the elimination equation. - Does elimination produce an identity or a contradiction?

1. Multiply equation (I) by \(4\): \(4ax - 36y = 24\). 2. Multiply equation (II) by \(a\): \(4ax - a^2y = 4a\). 3. Subtract the first transformed equation from the second: \((36 - a^2)y = 4a - 24\). 4. The system can fail to have exactly one solution only when \(36 - a^2 = 0\). Thus, \(a = 6\) or \(a = -6\). 5. If \(a = 6\), the elimination equation is \(0 = 0\), and the original equations are equivalent. The system has infinitely many solutions. 6. If \(a = -6\), the elimination equation is \(0 = -48\), a contradiction. The system has no solution.

The system has no solution for \(a = -6\) and infinitely many solutions for \(a = 6\).

- Compare all three coefficients after scaling one equation. - If the equations describe one line, a single ordered pair cannot list every solution. - Choose one coordinate freely and express the other in terms of it.

1. Multiplying equation (II) by \(-2\) gives \(4x-6y=10\), exactly equation (I). 2. The equations represent the same line, so the system has infinitely many solutions. 3. Let \(x=t\). From \(2x-3y=5\), \(y=\frac{2t-5}{3}\).

The system has infinitely many solutions: \(\left\{\left(t, \frac{2t-5}{3}\right)\mid t\in\mathbb{R}\right\}\).

- Substitute the parametric coordinates into the equation. - Then ask whether every point satisfying the equation can be expressed with a parameter. - Two different representations can still describe exactly the same set.

1. For a point \((t, 4-2t)\), the expression \(2x+y\) equals \(2t+(4-2t)=4\). 2. Thus, every parametrically described point satisfies \(2x+y=4\). 3. Conversely, any point on \(2x+y=4\) can be written as \((t, 4-2t)\) by taking \(t=x\). 4. The descriptions represent the same line.

The system has infinitely many solutions.

- Use the common nonzero ratio to compare the variable-coefficient pairs. - The constants determine whether the lines occupy the same position. - One entire equation must scale into the other for the lines to coincide.

1. The equal nonzero coefficient ratios mean \((a, b)=r(c, d)\), so the lines have the same slope. 2. If \(\frac{p}{q}=r\), one entire equation is a nonzero multiple of the other, so the system has infinitely many solutions. 3. If \(\frac{p}{q}\ne r\), the lines are distinct and parallel, so the system has no solution. 4. Exactly one solution is impossible.

No solution or infinitely many solutions are possible. Comparing \(\frac{p}{q}\) with \(r\) decides which case occurs.

- Determine how a half-turn changes a line’s direction. - Compare the case when the center lies on the line with the stated case. - Distinct parallel lines have no common point.

1. A \(180^\circ\) rotation preserves the direction of a line, so \(M\) is parallel to \(L\). 2. If \(M\) were the same line, the rotation center would lie midway between corresponding points of that line and would lie on the line. 3. Since \(P\) is not on \(L\), the image is a distinct parallel line, so the system has no solution.

The system has no solution.