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Algebra II : Solving Equations

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Example Questions

Example Question #31 : Solving Equations

Solve f(-14) if $f(x)=-0.5x+\frac{42}{x}$ .

Possible Answers:

Correct answer:

Explanation:

Plug in for in the f(x) equation to get

$f(-14)=7+(\frac{42}{-14})=7-3=4$

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Example Question #31 : How To Find F(X)

Solve for f(x) . When x=2 .

f(x)=5x^3+2x-24

Possible Answers:

f(x)=10

f(x)=-20

f(x)=14

f(x)=-15

f(x)=20

Correct answer:

f(x)=20

Explanation:

Given the equation,

f(x)=5x^3+2x-24 and x=2

Plug in for to the equation, f(2)=5(2)^3+2(2)-24

Solve and simplify.

f(2)=5(8)+4-24

f(2)=40+4-24

f(2)=44-24

f(2)=20

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Example Question #32 : How To Find F(X)

$f(x)=\frac{4x-2}{x+3}, x\neq-3$

Solve for f(x) , when x=0 .

Possible Answers:

$f(x)=\frac{1}{3}$

$f(x)=\frac{-1}{3}$

f(x)=0

$f(x)=\frac{-2}{3}$

$f(x)=\frac{2}{3}$

Correct answer:

$f(x)=\frac{-2}{3}$

Explanation:

$f(x)=\frac{4x-2}{x+3}, x\neq-3$

Plug in the value for .

$f(0)=\frac{4(0)-2}{(0)+3}$

Simplify

$f(0)=\frac{0-2}{3}$

Subtract

$f(0)=\frac{-2}{3}$

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Example Question #3 : Linear Systems With Two Variables

x+y=32

and

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

Solve for and .

Possible Answers:

$y=11\frac{2}{5},x=20\frac{3}{5}$

$y=19\frac{2}{5},x=12\frac{3}{5}$

$y=20\frac{2}{5},x=11\frac{3}{5}$

None of the available answers

$y=20\frac{3}{5},x=11\frac{4}{5}$

Correct answer:

$y=20\frac{2}{5},x=11\frac{3}{5}$

Explanation:

x+y=32 rearranges to

x=32-y

and

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

$32-y=\frac{2}{3}y-2$

$32-1\frac{2}{3}y=-2$

$-1\frac{2}{3}y=-34$

$-\frac{5}{3}y=-34$

$y=\frac{-34}{1}\cdot\frac{-3}{5}=\frac{102}{5}=20\frac{2}{5}$

$x=32-20\frac{2}{5}=11\frac{3}{5}$

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Example Question #4 : Linear Systems With Two Variables

Solve for (x, y) in the system of equations:

y = 3x + 4

2x + 3y = 34

Possible Answers:

(2, 10)

(3, 13)

(5, 19)

The system has no solution

(4, 16)

Correct answer:

(2, 10)

Explanation:

In the second equation, you can substitute 3x + 4 for from the first.

2x + 3y = 34

2x + 3 (3x + 4) = 34

2x + 3 (3x) + 3 (4) = 34

2x + 9x + 12 = 34

11x + 12 = 34

11x = 22

x = 2

Now, substitute 2 for in the first equation:

y = 3x + 4

y = 3 (2) + 4

y = 6 + 4

y = 10

The solution is (2, 10)

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Example Question #23 : How To Find The Solution To An Equation

Solve for :

ax-b=cd

Possible Answers:

$\frac{b}{cd+a}$

$\frac{cd+a}{b}$

$\frac{a}{cd+b}$

None of the other answers

$\frac{cd+b}{a}$

Correct answer:

$\frac{cd+b}{a}$

Explanation:

To solve for , you must isolate it from the other variables. Start by adding to both sides to give you ax=cd+b . Now, you need only to divide from both sides to completely isolate . This gives you a solution of $x=\frac{cd+b}{a}$ .

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Example Question #32 : Solving Equations

For the following equation, if x = 2, what is y?

y=x^3-4x+9

Possible Answers:

Correct answer:

Explanation:

On the equation, replace x with 2 and then simplify.

y=x^3-4x+9

y=(2)^3-4(2)+9

y=8-8+9

y=9

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Example Question #8 : How To Factor An Equation

Solve for $\small x$ .

x^2-5x-36

Possible Answers:

$x=0,\ 3$

$x=-2,\ 5$

$x=2,\ 12$

$x=-6,\ 4$

$x=-4,\ 9$

Correct answer:

$x=-4,\ 9$

Explanation:

x^2-5x-36

This is a quadratic equation. We can solve for $\small x$ either by factoring or using the quadratic formula. Since this equation is factorable, I will present the factoring method here.

The factored form of our equation should be in the format (A+B)(C+D)=0 .

To yield the first value in our original equation ( $\small x^2$ ), $\small A=x$ and $\small C=x$ .

(x+B)(x+D)=0

To yield the final term in our original equation ( $\small -36$ ), we can set $\small B=-9$ and $\small D=4$ .

$-9*4=-36\ \text{and}\ -9+4=-5$

(x-9)(x+4)=0

Now that the equation has been factored, we can evaluate $\small x$ . We set each factored term equal to zero and solve.

$\begin{matrix} x-9=0 &x+4=0 \\ x-9+9=0+9&x+4-4=0-4 \\ x=9 & x=-4 \end{matrix}$

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Example Question #31 : Equations / Solution Sets

Without drawing a graph of either equation, find the point where the two lines intersect.

Line 1 : y = 3x

Line 2 : y = x - 2

Possible Answers:

(1,3)

(0,0)

(2,0)

(0,-2)

(-1,-3)

Correct answer:

(-1,-3)

Explanation:

To find the point where these two lines intersect, set the equations equal to each other, such that is substituted with the side of the second equation. Solving this new equation for will give the -coordinate of the point of intersection.

3x = x - 2

Subtract from both sides.

(3x) - x = (x - 2) - x

2x = - 2

Divide both sides by 2.

$\frac{2x}{2} = \frac{-2}{2}$

x = - 1

Now substitute into either equation to find the -coordinate of the point of intersection.

y = 3x

y= 3(-1)

y = -3

With both coordinates, we know the point of intersection is (-1,-3) . One can plug in for and for in both equations to verify that this is correct.

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Example Question #32 : Equations / Solution Sets

What is the sum of and for the following system of equations?

3x - 5y = 5

-2x + 5y = 0

Possible Answers:

Correct answer:

Explanation:

3x - 5y = 5

-2x + 5y = 0

Add the equations together.

3x+(-2x)=x

-5y+5y=0

5+0=5

Put the terms together to see that $x+0=5\ \textup{or}\ x=5$ .

Substitute this value into one of the original equaitons and solve for .

3x - 5y = 5

3(5) - 5y = 5

-5y=-10

y=2

Now we know that $x=5\ \textup{and}\ y=2$ , thus we can find the sum of and .

x+y=5+2=7

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Algebra II : Solving Equations

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #31 : Solving Equations

Example Question #31 : How To Find F(X)

Example Question #32 : How To Find F(X)

Example Question #3 : Linear Systems With Two Variables

Example Question #4 : Linear Systems With Two Variables

Example Question #23 : How To Find The Solution To An Equation

Example Question #32 : Solving Equations

Example Question #8 : How To Factor An Equation

Example Question #31 : Equations / Solution Sets

Example Question #32 : Equations / Solution Sets

All Algebra II Resources

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