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Algebra II : Solving Equations

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Example Questions

Example Question #13 : Equations / Solution Sets

Solve this system of equations.

7x+4y+9z=0

8x-3y-6z=12

2x-y-3z=5

Possible Answers:

x=-1 , y=2 , $z=-\frac{5}{3}$

x=-2 , y=-2 , z=2

$x=\frac{1}{2}$ , $y=\frac{3}{2}$ , z=-3

x=1 , y=2 , $z=-\frac{5}{3}$

x=5 , y=-9 , z=3

Correct answer:

x=1 , y=2 , $z=-\frac{5}{3}$

Explanation:

Equation 1: 7x+4y+9z=0

Equation 2: 8x-3y-6z=12

Equation 3: 2x-y-3z=5

Adding the terms of the first and second equations together will yield 15x+y+3z=12 .

Then, add that to the third equation so that the y and z terms are eliminated. You will get 17x=17 .

This tells us that x = 1. Plug this x = 1 back into the systems of equations.

7+4y+9z=0

8-3y-6z=12

2-y-3z=5

Now, we can do the rest of the problem by using the substitution method. We'll take the third equation and use it to solve for y.

2-y-3z=5

-y=3z+5-2

y=-3z-3

Plug this y-equation into the first equation (or second equation; it doesn't matter) to solve for z.

7+4y+9z=0

7+4(-3z-3) +9z=0

7-12z-12+9z=0

-5-3z=0

$z=-\frac{5}{3}$

We can use this z value to find y

y=-3z-3

$y=-3\left ( -\frac{5}{3}\right )-3$

y=2

So the solution set is x = 1, y = 2, and z = –5/3.

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Example Question #1 : Solving Equations

Solve for : $x^{2}-144=0$

Possible Answers:

x=12,-12

x=10,-14

x=8,-4

x=14,-14

Correct answer:

x=12,-12

Explanation:

To solve this problem we can first add 144 to each side of the equation yielding $x^{2}=144$

Then we take the square root of both sides to get $\sqrt{x^{2}}=\sqrt{144}$

Then we calculate the square root of 144 which is $x=\pm12$ .

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Example Question #1 : Solving Equations

Solve this system of equations for :

2x + y = 12

x - 3y = 13

Possible Answers:

x = 4

x = 1

x = 9

x = 11

x = 7

Correct answer:

x = 7

Explanation:

Multiply the top equation by 3 on both sides, then add the second equation to eliminate the terms:

2x + y = 12

$3\cdot \left (2x + y \right )= 3\cdot 12$

6x+3y = 36

$\underline{x - 3y = 13}$

$7x\;\;\;\; \; =49$

$7x \div 7 =49 \div 7$

x = 7

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Example Question #2 : How To Factor A Variable

Solve for .

$y-7=\frac{4x+6y}{3}$

Possible Answers:

$y=-\frac{2}{3}x-7$

$y=\frac{2}{3}x-7$

$y=\frac{4}{3}x-7$

$y=-\frac{4}{3}x-7$

y=-{3}x-7

Correct answer:

$y=-\frac{4}{3}x-7$

Explanation:

$y-7=\frac{4x+6y}{3}$

Multiply both sides by 3:

$\frac{3}{1}(y-7)=(\frac{4x+6y}{3})\frac{3}{1}$

3(y-7)={4x+6y}

Distribute:

3y-21={4x+6y}

Subtract from both sides:

-21=4x+6y-3y

Add the terms together, and subtract from both sides:

-4x-21=3y

Divide both sides by :

$\frac{-4x-21=3y}{3}$

Simplify:

$y=-\frac{4}{3}x-7$

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Example Question #2321 : Algebra Ii

Solve for :

x(x - 2) = x^2 -8

Possible Answers:

Correct answer:

Explanation:

Distribute the x through the parentheses:

x² –2x = x² – 8

Subtract x² from both sides:

–2x = –8

Divide both sides by –2:

x = 4

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Example Question #3 : Solving Equations

Solve for :.

18x^3 - 8x = 0

Possible Answers:

$\left \{ - \frac{3}{2}, \frac{3}{2} \right\}$

$\left \{-\frac{2}{3}, 0, \frac{2}{3} \right \}$

$\left \{ \frac{3}{2}, 0, -\frac{3}{2} \right\}$

$\left \{-\frac{2}{3}, \frac{2}{3} \right \}$

Correct answer:

$\left \{-\frac{2}{3}, 0, \frac{2}{3} \right \}$

Explanation:

First factor the expression by pulling out :

18x^3 - 8x = 0

2x(9x^2-4) = 0

Factor the expression in parentheses by recognizing that it is a difference of squares:

2x(3x+2)(3x-2)= 0

Set each term equal to 0 and solve for the x values:

$2x=0 \rightarrow x=0$

$3x+2=0\rightarrow x=-\frac{2}{3}$

$3x-2=0\rightarrow x=\frac{2}{3}$

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Example Question #1 : Systems Of Equations

Solve the system of equations.

4x+y=16

2x+3y=18

Possible Answers:

(-3,4)

(-4,3)

(3,4)

None of the other answers are correct.

(4,3)

Correct answer:

(3,4)

Explanation:

Isolate in the first equation.

y=16-4x

Plug into the second equation to solve for .

2x+3(16-4x)=18

2x+48-12x=18

-10x+48=30

-10x=-30

x=3

Plug into the first equation to solve for .

4(3)+y=16

12+y=16

y=4

Now we have both the and values and can express them as a point: (3,4) .

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Example Question #1 : Linear Systems With Two Variables

Solve for and .

3x+2y=2

2x+2y=4

Possible Answers:

Cannot be determined.

$x=-2\ and\ y=4$

$x=-2\ and\ y=8$

$x=2\ and\ y=-8$

$x=2\ and\ y=-4$

Correct answer:

$x=-2\ and\ y=4$

Explanation:

1st equation: 3x+2y=2

2nd equation: 2x+2y=4

Subtract the 2nd equation from the 1st equation to eliminate the "2y" from both equations and get an answer for x:

Plug the value of into either equation and solve for :

2(-2)+2y=4

-4+2y=4

2y=4+4

2y=8

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Example Question #1 : Linear Systems With Three Variables

What is a solution to this system of equations:

$\left\{\begin{matrix} x+2y-z=6\\ 2x+y+z=9\\ 3x+4y+z=18\end{matrix}\right.$

Possible Answers:

$\left\{\begin{matrix} x=3\\ y=-2\\ z=-1\end{matrix}\right.$

$\left\{\begin{matrix} x=3\\ y=2\\ z=1\end{matrix}\right.$

$\left\{\begin{matrix} x=-3\\ y=2\\ z=1\end{matrix}\right.$

$\left\{\begin{matrix} x=3\\ y=-2\\ z=1\end{matrix}\right.$

$\left\{\begin{matrix} x=-3\\ y=2\\ z=-1\end{matrix}\right.$

Correct answer:

$\left\{\begin{matrix} x=3\\ y=2\\ z=1\end{matrix}\right.$

Explanation:

Step 1: Multiply first equation by −2 and add the result to the second equation. The result is:

\left\{\begin{matrix} x+2y-z=6\\ -3y+3z=-3\\ 3x+4y+z=18\end{matrix}\right.

Step 2: Multiply first equation by −3 and add the result to the third equation. The result is:

\left\{\begin{matrix} x+2y-z=6\\ -3y+3z=-3\\ -2y+4z=0\end{matrix}\right.

Step 3: Multiply second equation by −23 and add the result to the third equation. The result is:

\left\{\begin{matrix} x+2y-z=6\\ -3y+3z=-3\\ 2z=2\end{matrix}\right.

Step 4: solve for z .

Step 5: solve for y .

-3y+3\times1=-3

Step 6: solve for x by substituting y=2 and z=1 into the first equation.

$x+2\times2-1=6$

x=3

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Example Question #2 : Systems Of Equations

What is a solution to this system of equations?

$\left\{\begin{matrix} 2y=3x+11\\ y=-5x-14\end{matrix}\right.$

Possible Answers:

(1,3)

(-3,1)

(3,-1)

(-3,-1)

(3,1)

Correct answer:

(-3,1)

Explanation:

Substitute equation 2. into equation 1.,

$2\left(-5x-14\right)=3x+11$

-10x-28=3x+11

-13x=39

so, x=-3

Substitute x=-3 into equation 2:

$y=-5\cdot (-3)-14=1$

so, the solution is (-3,1) .

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Algebra II : Solving Equations

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #13 : Equations / Solution Sets

Example Question #1 : Solving Equations

Example Question #1 : Solving Equations

Example Question #2 : How To Factor A Variable

Example Question #2321 : Algebra Ii

Example Question #3 : Solving Equations

Example Question #1 : Systems Of Equations

Example Question #1 : Linear Systems With Two Variables

Example Question #1 : Linear Systems With Three Variables

Example Question #2 : Systems Of Equations

All Algebra II Resources

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