How to find the solution for a system of equations - Algebra 1

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Question

Solve for .

Answer

For the second equation, solve for in terms of .

Plug this value of y into the first equation.

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Question

Solve the following system of equations:

Answer

Solve the second equation for y:

x - 2y = 4

-2y = 4 - x

y = -2 + x/2

Plug this into the first equation:

3x + 2(-2 + x/2) = 8

Solve for x:

3x - 4 + x = 8

4x = 12

x = 3

Plug this into the second equation to get a value for y:

3 - 2y = 4

2y = -1

y = -0.5

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Question

Solve the following system of equations:

Answer

Set the two equations equal to one another:

2x - 2 = 3x + 6

Solve for x:

x = -8

Plug this value of x into either equation to solve for y. We'll use the top equation, but either will work.

y = 2 * (-8) - 2

y = -18

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Question

A cube has a volume of $cm^{3}$ . If its width is $3x^{2}$ , its length is , and its height is $x^{3}$ , find .

Answer

Since the object in question is a cube, each of its sides must be the same length. Therefore, to get a volume of $cm^{3}$ , each side must be equal to the cube root of , which is cm.

We can then set each expression equal to .

The first expression can be solved by either or , but the other two expressions make it evident that the solution is .

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Question

Solve the system for and .

Answer

The most simple method for solving systems of equations is to transform one of the equations so it allows for the canceling out of a variable. In this case, we can multiply by to get .

Then, we can add to this equation to yield , so .

We can plug that value into either of the original equations; for example, .

So, as well.

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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 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

What is the solution to the following system of equations:

Answer

By solving one equation for , and replacing in the other equation with that expression, you generate an equation of only 1 variable which can be readily solved.

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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

Give the solution to the system of equations below.

Answer

Solve the second equation for , allowing us to solve using the substitution method.

Substitute for in the first equation, and solve for .

Now, substitute for in either equation; we will choose the second. This allows us to solve for .

$2\cdot 5+y=13$

Now we can write the solution in the notation , or .

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Question

Find the solution to the following system of equations.

Answer

To solve this system of equations, use substitution. First, convert the second equation to isolate $\small a$ .

$a-2b = 0\rightarrow a=2b$

Then, substitute $\small 2b$ into the first equation for $\small a$ .

Combine terms and solve for $\small b$ .

Now that we know the value of $\small b$ , we can solve for $\small a$ using our previous substitution equation.

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Question

Two lines have equations of and . At what point do these lines intersect?

Answer

We can solve this problem by setting up a simple system of equations. First, we want to change the equations so one variable can cancel out. Multiplying the first equation by 2 and the second equation by 3 gives us a new system of and . These equations add up to or . Plugging in 7 for in either of the original two equations shows us that is equal to 1 and the point is .

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Question

Find the solution:

Answer

To solve this system of equations, we must first eliminate one of the variables. We will begin by eliminating the variables by finding the least common multiple of the variable's coefficients. The least common multiple of 3 and 2 is 6, so we will multiply each equation in the system by the corresponding number, like

.

By using the distributive property, we will end up with

Now, add down each column so that you have

$0x+13y=39 \rightarrow 13y=39$

Then you solve for and determine that .

But you're not done yet! To find , you have to plug your answer for back into one of the original equations:

Solve, and you will find that .

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Question

Find a solution for the following system of equations:

$\left{\begin{matrix} x-2y-1=3\ -x+2y+2=5 \end{matrix}\right.$

Answer

When we add the two equations, the and variables cancel leaving us with:

which means there is no solution for this system.

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Question

Solve the following system of equations:

$\left{\begin{matrix} 4x-2y=-3\ 4x+2y=-5 \end{matrix}\right.$

Answer

When we add the two equations, the variables cancel leaving us with:

Solving for we get:

$\frac{8x}{8}=\frac{-8}{8}$

We can then substitute our value for into one of the original equations and solve for :

$\begin{matrix} -4-2y=-3\ -4+4-2y=-3+4\ -2y=1\ y=-\frac{1}{2}\ \end{matrix}$

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Question

Solve this system of equations for :

Answer

Multiply the bottom equation by 5, then add to the top equation:

$5 \left (6x-y \right ) =5\cdot 24$

$\underline{\textrm{; } 3x + 5y = ; ; 1}$

$x = \frac{121}{33} = \frac{11}{3} = 3 \frac{2}{3}$

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Question

Solve this system of equations for :

Answer

Multiply the top equation by :

$-2 \cdot \left (3x + 5y \right )= -2 \cdot 1$

$-6x -10y \right )= -2$

Now add:

$\underline{-6x -10y= -2} \right )$

$-11y \div (-11) = 22\div (-11)$

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Question

Solve this system of equations for :

Answer

Multiply the bottom equation by , then add to the top equation:

$5 \left (6x-y \right ) =5\cdot (-9)$

$\underline{\textrm{; } 3x + 5y = ; ; 23}$

Divide both sides by

$\frac{33x}{33}=-\frac{22}{33}$

$\frac{22}{33}=-\frac{2}{3}$

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Question

Solve this system of equations for :

Answer

Multiply the top equation by :

$-2 \cdot \left (3x + 5y \right )= -2 \cdot 23$

$-6x -10y \right )= -46$

Now add:

$\underline{-6x -10y = -46}$

$-11y \div (-11) = -55\div (-11)$

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