Algebra II : Solving Equations

Study concepts, example questions & explanations for Algebra II

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Example Questions

Example Question #31 : Solving Equations

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

Possible Answers:

\displaystyle -1

\displaystyle 0

\displaystyle 4

\displaystyle 7

Correct answer:

\displaystyle 4

Explanation:

Plug in \displaystyle -14 for \displaystyle x in the \displaystyle f(x) equation to get

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

Example Question #32 : Solving Equations

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

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

Possible Answers:

\displaystyle f(x)=20

\displaystyle f(x)=14

\displaystyle f(x)=10

\displaystyle f(x)=-15

\displaystyle f(x)=-20

Correct answer:

\displaystyle f(x)=20

Explanation:

Given the equation,

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

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


Solve and simplify. 


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

\displaystyle f(2)=40+4-24

\displaystyle f(2)=44-24

\displaystyle f(2)=20

Example Question #85 : Functions And Lines

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

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

Possible Answers:

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

\displaystyle f(x)=0

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

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

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

Correct answer:

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

Explanation:

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

Plug in the \displaystyle x value for \displaystyle x.

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

Simplify

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

Subtract

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

Example Question #31 : Solving Equations

If

\displaystyle x+y=32

and

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

Solve for \displaystyle x and \displaystyle y.

Possible Answers:

None of the available answers

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

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

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

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

Correct answer:

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

Explanation:

\displaystyle x+y=32 rearranges to

\displaystyle x=32-y

and

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

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

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

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

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

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

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

Example Question #32 : Solving Equations

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

\displaystyle y = 3x + 4

\displaystyle 2x + 3y = 34

Possible Answers:

\displaystyle (3, 13)

\displaystyle (4, 16)

\displaystyle (2, 10)

The system has no solution

\displaystyle (5, 19)

Correct answer:

\displaystyle (2, 10)

Explanation:

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

\displaystyle 2x + 3y = 34

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

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

\displaystyle 2x + 9x + 12 = 34

\displaystyle 11x + 12 = 34

\displaystyle 11x = 22

\displaystyle x = 2

Now, substitute 2 for \displaystyle x in the first equation:

\displaystyle y = 3x + 4

\displaystyle y = 3 (2) + 4

\displaystyle y = 6 + 4 

\displaystyle y = 10

The solution is \displaystyle (2, 10)

Example Question #33 : Solving Equations

Solve for \displaystyle x:

\displaystyle ax-b=cd

Possible Answers:

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

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

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

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

None of the other answers

Correct answer:

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

Explanation:

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

Example Question #81 : Functions And Lines

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

\displaystyle y=x^3-4x+9

Possible Answers:

9

7

25

16

1

Correct answer:

9

Explanation:

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

\displaystyle y=x^3-4x+9

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

\displaystyle y=8-8+9

\displaystyle y=9

Example Question #3 : Equations / Solution Sets

Solve for \displaystyle \small x.

\displaystyle x^2-5x-36

Possible Answers:

\displaystyle x=2,\ 12

\displaystyle x=-2,\ 5

\displaystyle x=-4,\ 9

\displaystyle x=0,\ 3

\displaystyle x=-6,\ 4

Correct answer:

\displaystyle x=-4,\ 9

Explanation:

\displaystyle x^2-5x-36

This is a quadratic equation. We can solve  for \displaystyle \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 \displaystyle (A+B)(C+D)=0.

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

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

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

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

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

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

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

Example Question #34 : Solving Equations

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

Line 1 : \displaystyle y = 3x

Line 2 : \displaystyle y = x - 2

Possible Answers:

\displaystyle (0,0)

\displaystyle (1,3)

\displaystyle (-1,-3)

\displaystyle (0,-2)

\displaystyle (2,0)

Correct answer:

\displaystyle (-1,-3)

Explanation:

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

\displaystyle 3x = x - 2

Subtract \displaystyle x from both sides.

\displaystyle (3x) - x = (x - 2) - x

\displaystyle 2x = - 2

Divide both sides by 2.

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

\displaystyle x = - 1

Now substitute \displaystyle -1 into either equation to find the \displaystyle y-coordinate of the point of intersection.

\displaystyle y = 3x

\displaystyle y= 3(-1)

\displaystyle y = -3

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

Example Question #35 : Solving Equations

What is the sum of \displaystyle x and \displaystyle y for the following system of equations?

\displaystyle 3x - 5y = 5

\displaystyle -2x + 5y = 0

Possible Answers:

\displaystyle 5

\displaystyle 12

\displaystyle 9

\displaystyle 2

\displaystyle 7

Correct answer:

\displaystyle 7

Explanation:

\displaystyle 3x - 5y = 5

\displaystyle -2x + 5y = 0

Add the equations together.

\displaystyle 3x+(-2x)=x

\displaystyle -5y+5y=0

\displaystyle 5+0=5

Put the terms together to see that .

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

\displaystyle 3x - 5y = 5

\displaystyle 3(5) - 5y = 5

\displaystyle -5y=-10

\displaystyle y=2

Now we know that , thus we can find the sum of \displaystyle x and \displaystyle y.

\displaystyle x+y=5+2=7

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