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Systems of Linear Equations

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Question 1Question →

Consider the system of linear equations below.

3x−4y=72x+3y=16\begin{aligned} 3x - 4y &= 7 \\ 2x + 3y &= 16 \end{aligned}

If (x,y)(x, y) is the solution to the system of equations above, what is the value of x+yx + y?

Show answer & explanation

Answer: 7

Answer

7
The correct answer is 7. By multiplying the first equation by 3 and the second by 4, we get 9x−12y=219x - 12y = 21 and 8x+12y=648x + 12y = 64. Adding these equations eliminates the yy terms and gives 17x=8517x = 85, which simplifies to x=5x = 5. Substituting x=5x = 5 into the second equation yields 2(5)+3y=16  ⟹  10+3y=16  ⟹  3y=6  ⟹  y=22(5) + 3y = 16 \implies 10 + 3y = 16 \implies 3y = 6 \implies y = 2. The sum of the coordinates is x+y=5+2=7x + y = 5 + 2 = 7.

Step-by-Step Solution

1
Multiply the equations by constants to align the coefficients of one variable for elimination.
Multiplying the first equation by 3 gives 9x−12y=219x - 12y = 21. Multiplying the second equation by 4 gives 8x+12y=648x + 12y = 64.
This sets up the system so that the yy terms have opposite coefficients (−12-12 and +12+12), allowing them to be eliminated by addition.
2
Add the two modified equations to eliminate the yy variable and solve for xx.
(9x−12y)+(8x+12y)=21+64  ⟹  17x=85  ⟹  x=5(9x - 12y) + (8x + 12y) = 21 + 64 \implies 17x = 85 \implies x = 5.
Adding the equations eliminates yy and leaves a single linear equation in terms of xx.
3
Substitute the value of xx back into one of the original equations to solve for yy.
Substituting x=5x = 5 into the second equation 2x+3y=162x + 3y = 16 gives 2(5)+3y=16  ⟹  10+3y=16  ⟹  3y=6  ⟹  y=22(5) + 3y = 16 \implies 10 + 3y = 16 \implies 3y = 6 \implies y = 2.
Plugging the known variable value back in allows us to solve for the remaining unknown variable.
4
Calculate the value of the requested expression x+yx + y.
x+y=5+2=7x + y = 5 + 2 = 7.
The question specifically asks for the sum of xx and yy.

Key Concept

Solving systems of linear equations using the elimination method and evaluating linear combinations of the solutions.

Alternative Method

Alternatively, you can solve the first equation for xx in terms of yy: 3x=7+4y  ⟹  x=7+4y33x = 7 + 4y \implies x = \frac{7+4y}{3}. Substitute this expression into the second equation: 2(7+4y3)+3y=162\left(\frac{7+4y}{3}\right) + 3y = 16. Multiply the entire equation by 3 to clear the fraction: 2(7+4y)+9y=48  ⟹  14+8y+9y=48  ⟹  17y=34  ⟹  y=22(7+4y) + 9y = 48 \implies 14 + 8y + 9y = 48 \implies 17y = 34 \implies y = 2. Substitute y=2y = 2 back to find xx: x=7+4(2)3=5x = \frac{7+4(2)}{3} = 5. Thus, x+y=5+2=7x + y = 5 + 2 = 7.
Estimated Time:1m 30s
Question 2Question →

In the system of equations below, what is the value of yy?

4x+3y=254x + 3y = 25
2x+3y=172x + 3y = 17
Show answer & explanation

Answer: 3

Answer

The value of yy is 33.
Subtracting the second equation from the first equation gives (4x+3y)−(2x+3y)=25−17(4x + 3y) - (2x + 3y) = 25 - 17, which simplifies to 2x=82x = 8. Dividing by 2 gives x=4x = 4. Substituting x=4x = 4 into the second equation gives 2(4)+3y=172(4) + 3y = 17, which simplifies to 8+3y=178 + 3y = 17. Subtracting 8 from both sides gives 3y=93y = 9, and dividing by 3 gives y=3y = 3.

Step-by-Step Solution

1
Subtract the second equation from the first equation to eliminate the yy term.
2x=82x = 8
Subtracting the equations eliminates 3y3y since it is common to both equations, leaving a single variable equation.
2
Solve for xx by dividing both sides of the equation by 2.
x=4x = 4
Dividing isolates the variable xx so we can find its numerical value.
3
Substitute x=4x = 4 into the second equation 2x+3y=172x + 3y = 17 and solve for yy.
y=3y = 3
Substituting the value of xx leaves only the variable yy, which can then be isolated and solved.

Key Concept

Solving systems of linear equations using elimination
Question 3Question →

Two linear equations are defined as follows:

3x+2y=123x + 2y = 12
x−2y=4x - 2y = 4

If the ordered pair (x,y)(x, y) satisfies both equations, what is the value of xx?

Show answer & explanation

Answer: 4

Answer

The value of xx is 4.
Adding the two equations 3x+2y=123x + 2y = 12 and x−2y=4x - 2y = 4 eliminates the yy terms, resulting in 4x=164x = 16. Dividing both sides by 4 gives the value of xx as 4.

Step-by-Step Solution

1
Add the two equations together to eliminate the yy variable.
4x=164x = 16
Since the coefficients of yy are opposites (22 and −2-2), adding the equations eliminates yy directly.
2
Divide both sides of the equation by 4.
x=4x = 4
To isolate the variable xx.

Key Concept

Solving systems of linear equations using the elimination method.

Alternative Method

We can solve the second equation for xx to get x=2y+4x = 2y + 4. Substituting this expression into the first equation gives 3(2y+4)+2y=123(2y + 4) + 2y = 12, which simplifies to 6y+12+2y=126y + 12 + 2y = 12, or 8y=08y = 0, meaning y=0y = 0. Substituting y=0y = 0 back into x=2y+4x = 2y + 4 yields x=4x = 4.
Estimated Time:45s
Question 4Question →

Consider the system of equations below.

y=3x−4y = 3x - 4
x+y=8x + y = 8

What is the value of xx?

Show answer & explanation

Answer: 33

Answer

The value of xx is 33.
To solve the system, substitute the value of yy from the first equation into the second equation: x+(3x−4)=8x + (3x - 4) = 8. Combining the xx terms gives 4x−4=84x - 4 = 8. Adding 44 to both sides of the equation results in 4x=124x = 12. Dividing both sides by 44 isolates the variable, giving x=3x = 3.

Step-by-Step Solution

1
Substitute the expression for yy from the first equation into the second equation.
x+(3x−4)=8x + (3x - 4) = 8
This eliminates the variable yy to create a single equation in terms of xx.
2
Combine the like terms on the left side of the equation.
4x−4=84x - 4 = 8
Grouping the xx variables simplifies the expression.
3
Add 44 to both sides of the equation to isolate the term with xx.
4x=124x = 12
Performing the inverse operation of subtraction isolates the variable term.
4
Divide both sides of the equation by 44 to solve for xx.
x=3x = 3
Performing the inverse operation of multiplication isolates the variable xx.

Key Concept

Solving a system of linear equations using the substitution method.
Question 5Question →

If (x,y)(x, y) is the solution to the system of equations below, what is the value of xx?

y=2x+1y = 2x + 1
3x+2y=163x + 2y = 16
Show answer & explanation

Answer: 2

Answer

The correct answer is 2.
Substituting y=2x+1y = 2x + 1 into the second equation yields 3x+2(2x+1)=163x + 2(2x + 1) = 16. Expanding this expression gives 3x+4x+2=163x + 4x + 2 = 16, which simplifies to 7x+2=167x + 2 = 16. Subtracting 2 from both sides gives 7x=147x = 14, and dividing by 7 yields x=2x = 2.

Step-by-Step Solution

1
Substitute the expression for yy from the first equation into the second equation.
3x+2(2x+1)=163x + 2(2x + 1) = 16
This reduces the system of two variables to a single equation in terms of xx.
2
Expand and simplify the equation by distributing and combining like terms.
7x+2=167x + 2 = 16
Simplifying the equation makes it easier to isolate the variable xx.
3
Isolate the variable xx by performing inverse operations.
x=2x = 2
Subtracting 2 and then dividing by 7 isolates xx to find its value.

Key Concept

Solving a system of linear equations using the substitution method.

Alternative Method

Alternatively, you can multiply the first equation by 2 to get −4x+2y=2-4x + 2y = 2 and subtract it from 3x+2y=163x + 2y = 16 to eliminate yy, which also yields 7x=147x = 14 and thus x=2x = 2.
Estimated Time:1m 0s
Question 6Question →

In the system of linear equations shown below, what value of xx satisfies both equations?

y=2x−5y = 2x - 5
3x−2y=123x - 2y = 12
Show answer & explanation

Answer: −2-2

Answer

−2-2
To find the value of xx, substitute the expression for yy from the first equation, y=2x−5y = 2x - 5, into the second equation, 3x−2y=123x - 2y = 12. This yields 3x−2(2x−5)=123x - 2(2x - 5) = 12. Distributing the −2-2 across the terms in the parentheses results in 3x−4x+10=123x - 4x + 10 = 12. Simplifying the left side of the equation by combining like terms gives −x+10=12-x + 10 = 12. Subtracting 1010 from both sides results in −x=2-x = 2. Finally, multiplying or dividing both sides by −1-1 yields x=−2x = -2.

Step-by-Step Solution

1
Substitute the expression for yy from the first equation into the second equation
3x−2(2x−5)=123x - 2(2x - 5) = 12
To eliminate yy and obtain an equation containing only xx.
2
Distribute −2-2 to the terms inside the parentheses
3x−4x+10=123x - 4x + 10 = 12
To remove parentheses and prepare the equation for simplifying.
3
Combine like terms and solve for xx
−x+10=12-x + 10 = 12, which gives −x=2-x = 2, and thus x=−2x = -2
To isolate and find the value of xx.

Key Concept

Solving systems of linear equations using substitution
Question 7Question →
In a science experiment, the temperature yy, in degrees Celsius, and the time xx, in minutes, are related by a system of equations. The relationships are given by:
y=2x−3y = 2x - 3
3x−y=83x - y = 8
What is the value of xx in the solution to this system?
Show answer & explanation

Answer: 5

Answer

5
The correct answer is 5. Substituting the expression for yy from the first equation into the second equation yields 3x−(2x−3)=83x - (2x - 3) = 8. Distributing the negative sign results in 3x−2x+3=83x - 2x + 3 = 8. Combining the variable terms gives x+3=8x + 3 = 8. Subtracting 3 from both sides isolates xx, giving x=5x = 5.

Step-by-Step Solution

1
Substitute the expression for yy from the first equation, y=2x−3y = 2x - 3, into the second equation.
3x−(2x−3)=83x - (2x - 3) = 8
This eliminates the variable yy so the equation can be solved directly for xx.
2
Distribute the negative sign to both terms inside the parentheses.
3x−2x+3=83x - 2x + 3 = 8
Distributing the negative sign to both 2x2x and −3-3 simplifies the expression correctly.
3
Combine like terms on the left side of the equation.
x+3=8x + 3 = 8
Subtracting 2x2x from 3x3x leaves a single xx term.
4
Subtract 3 from both sides of the equation to isolate xx.
x=5x = 5
This isolates the variable xx to find the final solution.

Key Concept

Solving systems of linear equations using substitution
Estimated Time:1m 0s
Question 8Question →

Consider the system of equations below.

x=2yx = 2y
3x−y=103x - y = 10

If (x,y)(x, y) is the solution to the system of equations above, what is the value of xx?

Show answer & explanation

Answer: 4

Answer

The value of xx is 44.
The system of equations can be solved by substituting x=2yx = 2y from the first equation into the second equation, which gives 3(2y)−y=103(2y) - y = 10. Simplifying this equation yields 6y−y=106y - y = 10, or 5y=105y = 10. Dividing by 55 gives y=2y = 2. Substituting y=2y = 2 back into the first equation yields x=2(2)=4x = 2(2) = 4. Therefore, the value of xx is 44.

Step-by-Step Solution

1
Substitute the expression for xx from the first equation into the second equation.
3(2y)−y=103(2y) - y = 10
Since the first equation gives xx in terms of yy, substituting it into the second equation reduces the system to a single linear equation in one variable.
2
Solve the resulting equation for yy.
y=2y = 2
Simplifying 3(2y)−y=103(2y) - y = 10 yields 6y−y=106y - y = 10, which simplifies further to 5y=105y = 10. Dividing both sides by 55 gives y=2y = 2.
3
Substitute the value of yy back into the first equation to solve for xx.
x=4x = 4
Using x=2yx = 2y and substituting y=2y = 2 gives x=2(2)=4x = 2(2) = 4.

Key Concept

Solving systems of linear equations using the substitution method.
Question 9Question →

The variables xx and yy satisfy the system of equations below.

y=10−2xy = 10 - 2x
3x−y=53x - y = 5

What is the value of xx?

Show answer & explanation

Answer: 3

Answer

3
Substituting the expression for yy from the first equation, 10−2x10 - 2x, into the second equation yields 3x−(10−2x)=53x - (10 - 2x) = 5. Distributing the negative sign results in 3x−10+2x=53x - 10 + 2x = 5. Combining the like terms of xx gives 5x−10=55x - 10 = 5. Adding 10 to both sides of the equation results in 5x=155x = 15. Finally, dividing both sides by 5 gives x=3x = 3.

Step-by-Step Solution

1
Substitute the expression for yy from the first equation into the second equation.
3x−(10−2x)=53x - (10 - 2x) = 5
To eliminate the variable yy and obtain a single linear equation in terms of xx.
2
Distribute the negative sign through the parentheses and combine like terms.
3x−10+2x=53x - 10 + 2x = 5, which simplifies to 5x−10=55x - 10 = 5
To simplify the linear equation before isolating the variable.
3
Isolate the variable xx by adding 10 to both sides and then dividing by 5.
5x=155x = 15, which gives x=3x = 3
To determine the value of xx.

Key Concept

Solving systems of linear equations using substitution
Estimated Time:45s
Question 10Question →

In the system of equations below, what is the value of xx?

3x−y=73x - y = 7
x+y=5x + y = 5
Show answer & explanation

Answer: 3

Answer

The value of xx is 33.
Adding the two equations eliminates yy, resulting in 4x=124x = 12. Dividing by 44 gives x=3x = 3.

Step-by-Step Solution

1
Add the two equations to eliminate yy.
4x=124x = 12
Adding the equations (3x−y)+(x+y)=7+5(3x - y) + (x + y) = 7 + 5 eliminates yy because −y+y=0-y + y = 0.
2
Solve for xx.
x=3x = 3
Divide both sides of the equation 4x=124x = 12 by 44.

Key Concept

Solving a system of linear equations by elimination.
Question 11Question →

In the xyxy-plane, the graphs of the linear equations y=3x−5y = 3x - 5 and y=−x+7y = -x + 7 intersect at the point (x,y)(x, y). What is the value of xx?

Show answer & explanation

Answer: 3

Answer

3
To find the xx-value of the intersection point of the two graphs, set the two equations equal to each other because both represent the same yy-value at that point: 3x−5=−x+73x - 5 = -x + 7. Adding xx to both sides of the equation gives 4x−5=74x - 5 = 7. Adding 55 to both sides gives 4x=124x = 12. Dividing both sides by 44 yields x=3x = 3.

Step-by-Step Solution

1
Set the two expressions for yy equal to each other to find the xx-coordinate of the intersection point.
3x−5=−x+73x - 5 = -x + 7
Since both equations are solved for yy, their right-hand sides must be equal at the point of intersection.
2
Add xx to both sides of the equation to collect the variable terms on one side.
4x−5=74x - 5 = 7
Moving the variable terms together allows us to isolate xx.
3
Add 55 to both sides of the equation to isolate the term with xx.
4x=124x = 12
Grouping the constant terms on the opposite side prepares the equation for final division.
4
Divide both sides by 44 to find the value of xx.
x=3x = 3
This isolates xx and gives the final solution.

Key Concept

Solving a system of linear equations by setting the equations equal to find the point of intersection in the coordinate plane.
Question 12Question →

A food truck sells tacos and burritos. On Tuesday, the food truck sold a total of 120 tacos and burritos. The number of tacos sold, tt, was 20 more than the number of burritos sold, bb. This situation is represented by the system of equations below:

t+b=120t + b = 120
t=b+20t = b + 20

How many burritos did the food truck sell on Tuesday?

Show answer & explanation

Answer: 50

Answer

The food truck sold 50 burritos on Tuesday.
To find the number of burritos sold, bb, we substitute t=b+20t = b + 20 into the first equation t+b=120t + b = 120, giving (b+20)+b=120(b + 20) + b = 120. Combining like terms yields 2b+20=1202b + 20 = 120. Subtracting 20 from both sides results in 2b=1002b = 100. Finally, dividing both sides by 2 gives b=50b = 50. Thus, the food truck sold 50 burritos on Tuesday.

Step-by-Step Solution

1
Substitute the expression for tt from the second equation into the first equation
(b+20)+b=120(b + 20) + b = 120
To eliminate the variable tt and set up an equation with only one variable, bb.
2
Combine like terms on the left side of the equation
2b+20=1202b + 20 = 120
To simplify the equation for solving.
3
Subtract 20 from both sides of the equation
2b=1002b = 100
To isolate the variable term on one side of the equation.
4
Divide both sides of the equation by 2
b=50b = 50
To solve for the variable bb representing the number of burritos.

Key Concept

Solving systems of linear equations using the substitution method
Question 13Question →
A system of linear equations is shown below.
y=2x−7y = 2x - 7
3x−2y=113x - 2y = 11

What is the value of xx?

Show answer & explanation

Answer: 3

Answer

The correct value of xx is 3.
The correct value of xx is 3, which is found by substituting 2x−72x - 7 for yy in the second equation and solving for xx.

Step-by-Step Solution

1
Substitute the expression for yy from the first equation into the second equation.
3x−2(2x−7)=113x - 2(2x - 7) = 11
Since the first equation defines yy in terms of xx, substitution is a direct way to eliminate yy and solve for xx.
2
Distribute the −2-2 to both terms inside the parentheses.
3x−4x+14=113x - 4x + 14 = 11
Applying the distributive property requires multiplying −2-2 by both 2x2x and −7-7, remembering that negative times negative is positive.
3
Combine the like terms on the left side of the equation.
−x+14=11-x + 14 = 11
Simplifying the expression by combining the xx terms (3x−4x=−x3x - 4x = -x).
4
Isolate the variable term by subtracting 14 from both sides.
−x=−3-x = -3
Using the subtraction property of equality to move the constant term to the right side.
5
Solve for xx by multiplying or dividing both sides by −1-1.
x=3x = 3
Finding the positive value of xx.

Key Concept

Solving systems of linear equations using the substitution method.
Question 14Question →

If the equations x+2y=12x + 2y = 12 and x+y=7x + y = 7 are true for the same values of xx and yy, what is the value of yy?

Show answer & explanation

Answer: 5

Answer

The value of yy is 5.
Subtracting the equation x+y=7x + y = 7 from x+2y=12x + 2y = 12 gives (x−x)+(2y−y)=12−7(x - x) + (2y - y) = 12 - 7, which simplifies directly to y=5y = 5.

Step-by-Step Solution

1
Subtract the second equation from the first equation.
y=5y = 5
Subtracting x+y=7x + y = 7 from x+2y=12x + 2y = 12 eliminates the xx terms, leaving a single linear equation in terms of yy.

Key Concept

Solving systems of linear equations using elimination.
Question 15Question →

A system of two linear equations is given below.

y=x−2y = x - 2
2x−3y=12x - 3y = 1

If (x,y)(x, y) is the solution to the system of equations, what is the value of xx?

Show answer & explanation

Answer: 5

Answer

The value of xx is 5.
To find the value of xx, substitute the expression for yy from the first equation into the second equation: 2x−3(x−2)=12x - 3(x - 2) = 1. Distribute the −3-3 to obtain 2x−3x+6=12x - 3x + 6 = 1, which simplifies to −x+6=1-x + 6 = 1. Subtracting 6 from both sides gives −x=−5-x = -5. Multiplying or dividing both sides by −1-1 yields x=5x = 5.

Step-by-Step Solution

1
Substitute the expression for yy from the first equation into the second equation.
2x−3(x−2)=12x - 3(x - 2) = 1
Substituting the expression reduces the system of two equations to one linear equation containing only the variable xx.
2
Distribute the coefficient −3-3 to both terms inside the parentheses.
2x−3x+6=12x - 3x + 6 = 1
Applying the distributive property removes the parentheses so that like terms can be combined.
3
Combine like terms and isolate xx.
−x+6=1  ⟹  −x=−5  ⟹  x=5-x + 6 = 1 \implies -x = -5 \implies x = 5
Combining the xx terms and subtracting 6 from both sides isolates −x-x, allowing you to find the final value of xx.

Key Concept

Solving systems of linear equations using the substitution method

Alternative Method

Alternatively, you can solve this system using elimination. First, rewrite the first equation in standard form by subtracting xx from both sides, which gives −x+y=−2-x + y = -2. Next, multiply this equation by 3 to align coefficients: −3x+3y=−6-3x + 3y = -6. Adding this result directly to the second equation, 2x−3y=12x - 3y = 1, eliminates the yy terms and leaves −x=−5-x = -5, which simplifies to x=5x = 5.
Estimated Time:45s
Question 16Question →

A system of equations is shown below.

y=−2(x−5)y = -2(x - 5)
3x+y=133x + y = 13

What is the value of xx?

Show answer & explanation

Answer: 3

Answer

The value of xx is 33.
Substituting the expression for yy from the first equation into the second equation gives 3x−2(x−5)=133x - 2(x - 5) = 13. Distributing the −2-2 yields 3x−2x+10=133x - 2x + 10 = 13. Combining the xx terms gives x+10=13x + 10 = 13. Subtracting 10 from both sides of the equation yields x=3x = 3.

Step-by-Step Solution

1
Substitute the expression for yy from the first equation into the second equation.
3x−2(x−5)=133x - 2(x - 5) = 13
This substitution reduces the system to a single linear equation in terms of xx.
2
Apply the distributive property to simplify the equation.
3x−2x+10=133x - 2x + 10 = 13
Multiplying −2-2 by xx gives −2x-2x, and multiplying −2-2 by −5-5 gives +10+10.
3
Combine like terms and isolate xx.
x+10=13x + 10 = 13, which simplifies to x=3x = 3
Subtracting 10 from both sides gives the value of xx.

Key Concept

Solving a system of linear equations using substitution.
Estimated Time:1m 0s
Question 17Question →

In the system of equations below, kk is a constant.

kx−3y=4kx - 3y = 4
(k−2)x−5y=8(k-2)x - 5y = 8

If the system has no solution, what is the value of kk?

Show answer & explanation

Answer: -3

Answer

-3
To find the value of kk for which the system of linear equations has no solution, we determine when the two lines represented by the equations are parallel and distinct. Parallel lines have equal slopes. We can express each equation in slope-intercept form, y=mx+by = mx + b. For the first equation, kx−3y=4kx - 3y = 4 simplifies to y=k3x−43y = \frac{k}{3}x - \frac{4}{3}, giving a slope of k3\frac{k}{3}. For the second equation, (k−2)x−5y=8(k-2)x - 5y = 8 simplifies to y=k−25x−85y = \frac{k-2}{5}x - \frac{8}{5}, giving a slope of k−25\frac{k-2}{5}. Equating the two slopes yields k3=k−25\frac{k}{3} = \frac{k-2}{5}. Cross-multiplying gives 5k=3(k−2)5k = 3(k-2), which simplifies to 5k=3k−65k = 3k - 6. Subtracting 3k3k from both sides gives 2k=−62k = -6, and dividing by 22 results in k=−3k = -3. Since the y-intercepts −43-\frac{4}{3} and −85-\frac{8}{5} are distinct, the lines are parallel and do not intersect, meaning there is no solution.

Step-by-Step Solution

1
Express both equations in slope-intercept form (y=mx+by = mx + b) to find their slopes.
For the first equation, kx−3y=4  ⟹  3y=kx−4  ⟹  y=k3x−43kx - 3y = 4 \implies 3y = kx - 4 \implies y = \frac{k}{3}x - \frac{4}{3} (slope is k3\frac{k}{3}). For the second equation, (k−2)x−5y=8  ⟹  5y=(k−2)x−8  ⟹  y=k−25x−85(k-2)x - 5y = 8 \implies 5y = (k-2)x - 8 \implies y = \frac{k-2}{5}x - \frac{8}{5} (slope is k−25\frac{k-2}{5}).
A system of two linear equations has no solution if the lines are parallel (slopes are equal) and their y-intercepts are different.
2
Set the slopes equal to each other.
k3=k−25\frac{k}{3} = \frac{k-2}{5}
Since the lines must be parallel, their slopes must be equivalent.
3
Solve for kk by cross-multiplying.
5k=3(k−2)  ⟹  5k=3k−6  ⟹  2k=−6  ⟹  k=−35k = 3(k-2) \implies 5k = 3k - 6 \implies 2k = -6 \implies k = -3.
Cross-multiplication isolates the variable kk. Since the y-intercepts (−43-\frac{4}{3} and −85-\frac{8}{5}) are different, this value of kk guarantees the lines are parallel and distinct.

Key Concept

Determining parameters for a system of linear equations with no solution

Alternative Method

Alternatively, align the coefficients of yy to eliminate the yy terms. Multiply the first equation by 55 and the second equation by −3-3:

5(kx−3y)=5(4)  ⟹  5kx−15y=205(kx - 3y) = 5(4) \implies 5kx - 15y = 20
−3((k−2)x−5y)=−3(8)  ⟹  −3(k−2)x+15y=−24-3((k-2)x - 5y) = -3(8) \implies -3(k-2)x + 15y = -24

Adding these two equations yields:
(5k−3(k−2))x=−4(5k - 3(k-2))x = -4
(2k+6)x=−4(2k + 6)x = -4

For the system to have no solution, the coefficient of xx must be 00 while the constant on the right side is non-zero (which it is, −4-4). Setting the coefficient of xx to 00:
2k+6=0  ⟹  k=−32k + 6 = 0 \implies k = -3
Estimated Time:2m 0s
Question 18Question →
3x−5y=8kx+10y=−3\begin{aligned} 3x - 5y &= 8 \\ kx + 10y &= -3 \end{aligned}

In the system of equations above, kk is a constant. If the system has no solution, what is the value of kk?

Show answer & explanation

Answer: -6

Answer

-6
The correct answer is −6-6. A system of two linear equations has no solution when the equations represent parallel lines, meaning they have the same slope but different yy-intercepts. Writing the first equation in slope-intercept form gives y=35x−85y = \frac{3}{5}x - \frac{8}{5}, so its slope is 35\frac{3}{5}. Writing the second equation in slope-intercept form gives y=−k10x−310y = -\frac{k}{10}x - \frac{3}{10}, so its slope is −k10-\frac{k}{10}. Since the yy-intercepts are different (−85≠−310-\frac{8}{5} \neq -\frac{3}{10}), the system will have no solution when their slopes are equal: 35=−k10\frac{3}{5} = -\frac{k}{10}. Solving this equation for kk yields k=−6k = -6.

Step-by-Step Solution

1
Convert the first equation to slope-intercept form (y=mx+by = mx + b) to find its slope.
y=35x−85y = \frac{3}{5}x - \frac{8}{5}, so the slope is 35\frac{3}{5} and the yy-intercept is −85-\frac{8}{5}.
Expressing the line in slope-intercept form directly reveals its slope and yy-intercept.
2
Convert the second equation to slope-intercept form to express its slope in terms of kk.
10y=−kx−3⇒y=−k10x−31010y = -kx - 3 \Rightarrow y = -\frac{k}{10}x - \frac{3}{10}, so the slope is −k10-\frac{k}{10} and the yy-intercept is −310-\frac{3}{10}.
This allows comparison of the slope and yy-intercept of the second line with the first line.
3
Set the slopes equal to each other and solve for kk.
35=−k10⇒30=−5k⇒k=−6\frac{3}{5} = -\frac{k}{10} \Rightarrow 30 = -5k \Rightarrow k = -6.
A system of two linear equations has no solution if and only if the lines are parallel (equal slopes) and have different yy-intercepts (which −85-\frac{8}{5} and −310-\frac{3}{10} are).

Key Concept

Determining the number of solutions to a system of linear equations based on slope and intercept relationships.
Question 19Question →

In the system of equations below, aa is a constant.

ax+3y=52x−y=1\begin{aligned} ax + 3y &= 5 \\ 2x - y &= 1 \end{aligned}

If the system has a solution (x,y)(x, y) that lies in Quadrant IV of the xyxy-plane, which of the following could be the value of aa?

Show answer & explanation

Answer: 12

Answer

12
The correct answer is 12. For the solution (x,y)(x, y) to lie in Quadrant IV, the xx-coordinate must be positive and the yy-coordinate must be negative. Isolating yy in the second equation gives y=2x−1y = 2x - 1. Substituting this into the first equation yields x=8a+6x = \frac{8}{a+6}. For xx to be positive, we must have a>−6a > -6. Substituting xx back into the expression for yy gives y=10−aa+6y = \frac{10-a}{a+6}. For yy to be negative when a>−6a > -6, we must have 10−a<010 - a < 0, which simplifies to a>10a > 10. Among the choices, only 12 is greater than 10.

Step-by-Step Solution

1
Identify the signs of coordinates in Quadrant IV.
x>0x > 0 and y<0y < 0
In Quadrant IV of the coordinate plane, the horizontal coordinate is positive and the vertical coordinate is negative.
2
Express yy in terms of xx from the second equation.
y=2x−1y = 2x - 1
This facilitates substituting yy into the first equation to solve for xx.
3
Substitute y=2x−1y = 2x - 1 into the first equation and solve for xx.
x=8a+6x = \frac{8}{a + 6}
To find the coordinates of the intersection point in terms of the parameter aa.
4
Substitute x=8a+6x = \frac{8}{a + 6} back into the expression for yy.
y=10−aa+6y = \frac{10 - a}{a + 6}
To express the yy-coordinate of the intersection point in terms of the parameter aa.
5
Apply the conditions x>0x > 0 and y<0y < 0 to solve for aa.
a>10a > 10
For x>0x > 0, we need a+6>0  ⟹  a>−6a + 6 > 0 \implies a > -6. For y<0y < 0 given a+6>0a + 6 > 0, we need 10−a<0  ⟹  a>1010 - a < 0 \implies a > 10.
6
Evaluate the options to find a value of aa that satisfies the condition.
12
Among the options (12, 8, -4, -8), only 12 is greater than 10.

Key Concept

Solving systems of linear equations with parameters and applying quadrant constraints
Question 20Question →

In the system of equations below, aa and bb are constants.

3x−4y=7ax−12y=b\begin{aligned} 3x - 4y &= 7 \\ ax - 12y &= b \end{aligned}

If the system has infinitely many solutions, what is the value of a+ba + b?

Show answer & explanation

Answer: 30

Answer

30
For a system of two linear equations in two variables to have infinitely many solutions, the two equations must represent the same line. Comparing the yy-coefficients shows that −12-12 is 33 times −4-4. Multiplying the first equation by 33 gives the equivalent equation 9x−12y=219x - 12y = 21. For this to be identical to the second equation, ax−12y=bax - 12y = b, the corresponding coefficients must be equal, meaning a=9a = 9 and b=21b = 21. Therefore, the value of a+ba + b is 9+21=309 + 21 = 30.

Step-by-Step Solution

1
Identify the condition for a system of linear equations to have infinitely many solutions.
The two equations must represent the same line, making them equivalent equations.
Infinitely many solutions occur when the two equations in a system are scalar multiples of each other.
2
Align the coefficients of the yy-terms in both equations.
Multiplying the first equation by 33 gives 9x−12y=219x - 12y = 21.
The coefficient of yy in the second equation is −12-12, which is 33 times the coefficient of yy in the first equation (−4-4).
3
Equate the coefficients of the remaining terms to find the values of aa and bb.
a=9a = 9 and b=21b = 21.
For 9x−12y=219x - 12y = 21 and ax−12y=bax - 12y = b to be equivalent, their corresponding coefficients and constants must be equal.
4
Calculate the value of a+ba + b.
a+b=9+21=30a + b = 9 + 21 = 30.
The problem asks for the value of the sum a+ba + b.

Key Concept

Determining constants in a system of linear equations with infinitely many solutions
Estimated Time:1m 30s
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