Solving Equations - Algebra II

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Question

Solve for :

Answer

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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Question

Solve for . When .

Answer

Given the equation,

and

Plug in for to the equation,

Solve and simplify.

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Question

Solve for .

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

Answer

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

Multiply both sides by 3:

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

Distribute:

Subtract from both sides:

Add the terms together, and subtract from both sides:

Divide both sides by :

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

Simplify:

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

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Question

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

Solve for , when .

Answer

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

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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Question

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

Answer

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

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Question

Solve this system of equations.

Answer

Equation 1:

Equation 2:

Equation 3:

Adding the terms of the first and second equations together will yield .

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

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

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.

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

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

We can use this z value to find y

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

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

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Question

If

and

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

Solve for and .

Answer

rearranges to

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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Question

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

Answer

To solve this problem we can first add 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 which is $x=\pm12$ .

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Question

Solve for in the system of equations:

Answer

In the second equation, you can substitute for from the first.

Now, substitute 2 for in the first equation:

The solution is

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Question

Solve the system of equations.

Answer

Isolate in the first equation.

Plug into the second equation to solve for .

Plug into the first equation to solve for .

Now we have both the and values and can express them as a point: .

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Question

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

Answer

Add the equations together.

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 .

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

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Question

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

Line 1 :

Line 2 :

Answer

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.

Subtract from both sides.

Divide both sides by 2.

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

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

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

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Question

Solve for :.

Answer

First factor the expression by pulling out :

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

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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Question

Solve for :

Answer

To solve for , you must isolate it from the other variables. Start by adding to both sides to give you . 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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Question

Solve for :

Answer

First, you must multiply the left side of the equation using the distributive property.

This gives you .

Next, subtract from both sides to get .

Then, divide both sides by to get .

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Question

Solve for :

Answer

Combine like terms on the left side of the equation:

Use the distributive property to simplify the right side of the equation:

Next, move the 's to one side and the integers to the other side:

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Question

Solve for :

Answer

To solve for , you must isolate it so that all of the other variables are on the other side of the equation. To do this, first subtract from both sides to get . Then, divide both sides by to get $x=\frac{g-f}{e}$ .

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Question

Solve for $\small x$ .

Answer

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 .

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

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

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

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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Question

Solve this system of equations for :

Answer

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

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

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

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

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Question

Solve for and .

Answer

1st equation:

2nd equation:

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 :

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