Algebra 1 : Equations / Solution Sets

Example Questions

Example Question #18 : How To Find A Solution Set

Solve the following by substitution:

No solution

Explanation:

To solve this equation you will plug the second equation straight into the first one by substituting what is written there for .

You will then get:

From here you need to simplify by combining like terms:

Then divide both sides by  to get:

You will then take this value of  and plug it into either equation.

Example Question #19 : How To Find A Solution Set

Set A is composed of all multiples of 4 that are that are less than the square of 7. Set B includes all multiples of 6 that are greater than 0. How many numbers are found in both set A and set B?

Explanation:

Start by making a list of the multiples of 4 that are smaller than the square of 7. When 7 is squared, it equals 49; thus, we can compose the following list:

Next, make a list of all the multiples of 6 that are greater than 0. Since we are looking for shared multiples, stop after 48 because numbers greater than 48 will not be included in set A. The biggest multiple of 4 smaller that is less than 49 is 48; therefore, do not calculate multiples of 6 greater than 48.

Finally, count the number of multiples found in both sets. Both sets include the following numbers:

The correct answer is 4 numbers.

Example Question #1 : How To Find The Solution For A System Of Equations

A cube has a volume of  . If its width is  , its length is , and its height is  , find .

Explanation:

Since the object in question is a cube, each of its sides must be the same length. Therefore, to get a volume of  , 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 .

Example Question #2 : How To Find The Solution For A System Of Equations

Solve the system for  and .

Explanation:

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:

Explanation:

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 #4 : How To Find The Solution For A System Of Equations

Solve this system of equations for :

None of the other choices are correct.

Explanation:

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

Example Question #5 : How To Find The Solution For A System Of Equations

Solve this system of equations for :

None of the other choices are correct.

Explanation:

Multiply the top equation by :

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

Solve this system of equations for :

None of the other choices are correct.

Explanation:

Multiply the top equation by :

Example Question #7 : How To Find The Solution For A System Of Equations

Find the solution to the following system of equations.

Explanation:

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 #8 : How To Find The Solution For A System Of Equations

Find a solution for the following system of equations:

infinitely many solutions

no solution