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

## Example Questions

### Example Question #51 : How To Find The Solution For A System Of Equations

Find the solution to this system of equations:

None of these

Explanation:

To find the solution we will multiply by some number to cancel out one variable and then solve for the variable that is left over. This is called the "elimination" method. You may choose to eliminate either variable, but it is prudent to choose the variable that is easier to cancel.

1)

2)

Multiply equation 1 by -2:

Add new equation 1 to equation 2:

The result is

Solve for y:

Now solve for x by plugging in the value for y (we can choose either original equation):

### Example Question #52 : How To Find The Solution For A System Of Equations

Find the solution set for this system of equations:

None of these

Explanation:

To solve this system of equations, multiply each equation by an integer that would allow us to cancel one variable. We then substitute that value back into either original equation to find the other variable. In this case we will cancel the x variable by multiplying each equation by 3 and 4 respectively.

1)

2)

Multiply equation 1 by 3 and equation 2 by 4:

The result is

Solve for y:

Now substitute this value back into equation 1:

### Example Question #51 : How To Find The Solution For A System Of Equations

Give the solution to the system of equations below.

No solution

Explanation:

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 .

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

### Example Question #1 : Solve Systems Of Two Linear Equations: Ccss.Math.Content.8.Ee.C.8b

Solve this system of equations for :

None of the other choices are correct.

Explanation:

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

Divide both sides by

### Example Question #221 : Grade 8

Solve the following system of equations:

no solution

infinitely many solutions

Explanation:

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

Solving for  we get:

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

### Example Question #2451 : Algebra 1

Solve the following system of equations.

Explanation:

We are given

We can solve this by using the substitution method.  Notice that you can plug  from the first equation into the second equation and then get

Divide both sides by 5

So . We can use this value to find y by using either equation. In this case, I'll use .

So the solution is

### Example Question #1 : Solve Problems Leading To Two Linear Equations: Ccss.Math.Content.8.Ee.C.8c

Adult tickets to the zoo sell for ; child tickets sell for . On a given day, the zoo sold  tickets and raised  in admissions. How many adult tickets were sold?

Explanation:

Let  be the number of adult tickets sold. Then the number of child tickets sold is .

The amount of money raised from adult tickets is ; the amount of money raised from child tickets is . The sum of these money amounts is , so the amount of money raised can be defined by the following equation:

To find the number of adult tickets sold, solve for :

### Example Question #5 : Solve Problems Leading To Two Linear Equations: Ccss.Math.Content.8.Ee.C.8c

Solve the following story problem:

Jack and Aaron go to the sporting goods store. Jack buys a glove for  and  wiffle bats for  each. Jack has  left over. Aaron spends all his money on  hats for  each and  jerseys. Aaron started with  more than Jack. How much does one jersey cost?

Explanation:

Let's call "" the cost of one jersey (this is the value we want to find)

Let's call the amount of money Jack starts with ""

Let's call the amount of money Aaron starts with ""

We know Jack buys a glove for  and  bats for  each, and then has  left over after. Thus:

simplifying,  so Jack started with

We know Aaron buys  hats for  each and  jerseys (unknown cost "") and spends all his money.

The last important piece of information from the problem is Aaron starts with  dollars more than Jack. So:

From before we know:

Plugging in:

so Aaron started with

Finally we plug  into our original equation for A and solve for x:

Thus one jersey costs

### Example Question #2453 : Algebra 1

Solve for  only:

Explanation:

Use the elimination process to determine the value of .

Subtract the second equation with the first equation.  This will eliminate the x-variable in order to find the y-variable.

The equation becomes:

Use this value and substitute into either the first or second equation and solve for the x-variable.

Let's select the first equation and solve for .

The result will be the same if we substitute  into the second equation.

### Example Question #2461 : Algebra 1

Solve only for  given the two equations:

Explanation:

In order to solve for , multiply the first equation by four in order to eliminate the fractions in front of the coefficients.

The first equation becomes:

The second equation remains the same:

Subtract the second equation with the first equation:

The result is:

Divide by two on both sides.

Simplify both sides.