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Algebra II : Equations

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Example Questions

Example Question #2321 : Algebra Ii

Simplify the expression by combining like terms.

Simplify:

5x^2+15x+18 = 6x^2-8x+9

Possible Answers:

x^2+23x+9 = 0

-x^2-23x+27 = 0

5x^2-23x-9 = 6x^2

11x^2-23x-9 = 0

x^2-23x-9 = 0

Correct answer:

x^2-23x-9 = 0

Explanation:

The original equation:

5x^2+15x+18 = 6x^2-8x+9

Now move all terms to one side, in this example, we will use the right side, but either side will work.

0 = 6x^2 - 8x + 9 -5x^2 -15x-18

As you can see, we subtracted the terms on the left from both sides, effectively moving them to the other side.

Regrouping the terms so that "like terms" are together. Like terms are defined by having the same power of x.

0= (6x^2-5x^2)+(-8x-15x)+(9-18)

Now, we add any terms that have like powers of x.

x^2-23x-9 = 0

Now that all terms have been combined, we are finished. The equation is simplified.

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Example Question #1 : Linear Systems With Three Variables

Solve this system of equations.

7x+4y+9z=0

8x-3y-6z=12

2x-y-3z=5

Possible Answers:

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

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

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

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

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

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 #3 : How To Find A Solution Set

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

Possible Answers:

x=12,-12

x=10,-14

x=14,-14

x=8,-4

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 #2 : Solving Equations

Solve this system of equations for :

2x + y = 12

x - 3y = 13

Possible Answers:

x = 7

x = 1

x = 11

x = 9

x = 4

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:

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

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

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

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

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 #1 : Solving Equations

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 #4 : Solving Equations

Solve for :.

18x^3 - 8x = 0

Possible Answers:

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

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

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

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

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 : Linear Systems With Two Variables

Solve the system of equations.

4x+y=16

2x+3y=18

Possible Answers:

None of the other answers are correct.

(-3,4)

(4,3)

(-4,3)

(3,4)

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 : Systems Of Equations

Solve for and .

3x+2y=2

2x+2y=4

Possible Answers:

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

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

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

Cannot be determined.

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

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 #2 : 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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Algebra II : Equations

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #2321 : Algebra Ii

Example Question #1 : Linear Systems With Three Variables

Example Question #3 : How To Find A Solution Set

Example Question #2 : Solving Equations

Example Question #2 : How To Factor A Variable

Example Question #1 : Solving Equations

Example Question #4 : Solving Equations

Example Question #1 : Linear Systems With Two Variables

Example Question #1 : Systems Of Equations

Example Question #2 : Linear Systems With Three Variables

All Algebra II Resources

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