### All Common Core: 8th Grade Math Resources

## Example Questions

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

Solve the system for and .

**Possible Answers:**

**Correct 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.

### Example Question #3 : How To Find The Solution For A System Of Equations

What is the solution to the following system of equations:

**Possible Answers:**

**Correct 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.

### Example Question #2 : Solve Systems Of Two Linear Equations: Ccss.Math.Content.8.Ee.C.8b

Solve this system of equations for :

**Possible Answers:**

None of the other choices are correct.

**Correct answer:**

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

### Example Question #2 : Solve Systems Of Two Linear Equations: Ccss.Math.Content.8.Ee.C.8b

Solve this system of equations for :

**Possible Answers:**

None of the other choices are correct.

**Correct answer:**

Multiply the top equation by :

Now add:

### Example Question #3 : Solve Systems Of Two Linear Equations: Ccss.Math.Content.8.Ee.C.8b

Solve this system of equations for :

**Possible Answers:**

None of the other choices are correct.

**Correct answer:**

Multiply the top equation by :

Now add:

### Example Question #6 : How To Find The Solution For A System Of Equations

Find the solution to the following system of equations.

**Possible Answers:**

**Correct answer:**

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

Then, substitute into the first equation for .

Combine terms and solve for .

Now that we know the value of , we can solve for using our previous substitution equation.

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

Find a solution for the following system of equations:

**Possible Answers:**

no solution

infinitely many solutions

**Correct answer:**

no solution

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

which means there is no solution for this system.

### Example Question #5 : Solve Systems Of Two Linear Equations: Ccss.Math.Content.8.Ee.C.8b

Solve the set of equations:

**Possible Answers:**

**Correct answer:**

Solve the first equation for :

Substitute into the second equation:

Multiply the entire equation by 2 to eliminate the fraction:

Using the value of , solve for :

Therefore, the solution is

### Example Question #6 : Solve Systems Of Two Linear Equations: Ccss.Math.Content.8.Ee.C.8b

Solve the following system of equations:

**Possible Answers:**

**Correct 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

### Example Question #7 : Solve Systems Of Two Linear Equations: Ccss.Math.Content.8.Ee.C.8b

Solve this system of equations for :

**Possible Answers:**

None of the other choices are correct.

**Correct answer:**

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

Divide both sides by

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### All Common Core: 8th Grade Math Resources

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