Solving Two-Step Equations

Hello! Today we are going to take a look at solving two-step equations. So let’s start with an example:

\(3x-4=11\)

 

To solve, we need to find the value of \(x\) that makes this statement true. We do this by getting \(x\) by itself on one side of the equation.

When solving two-step equations, always undo the addition or subtraction first, and then undo the multiplication or division.

So, for this problem, we’re going to start by undoing the subtraction by adding 4 to both sides.

\(3x-4+4=11+4\)

 

That means these cancel out, and we’re left with \(3x\) on the left side, and \(11+4=15\).

\(3x=15\)

 

Then, we undo the 3 being multiplied by \(x\) by dividing both sides by 3. Remember, any time a number is next to a letter, it means they are being multiplied. So this shows us multiplication. So, to undo that, we simply divide both sides by 3.

\(\frac{3x}{3}=\frac{15}{3}\)

 

Our \(3s\) cancel out and we’re left with \(x\) on the left side, and \(15\div 3=5\).

\(x=5\)

 

And that’s our answer! We can check it by plugging it back into our original equation and seeing if it is a true statement. So let’s do that.

\(3(5)-4=11\)
 
\(15-4=11\)
 
\(11=11\)

 

\(11=11\) is a true statement, so that verifies that \(x=5\) is the correct answer.

Let’s take a look at another problem.

\(\frac{x}{4}+5=7\)

 

Remember, we first want to undo the addition. So, to undo addition by 5, we’re going to subtract 5 from both sides.

\(\frac{x}{4}+5-5=7-5\)

 

That gives us \(\frac{x}{4}\) (or \(x\div 4\)), these (5s) cancel out, remember, and then \(7-5=2\).

\(\frac{x}{4}=2\)

 

So, now we’re left with division: \(x\) is being divided by 4. So, to undo this, we simply multiply both sides by 4.

\(4\times \frac{x}{4}=24\)

 

Our 4s cancel out and we’re left with \(x\) on the left side, and \(2\times 4=8\).

\(x=8\)

 

So our answer is \(x=8\).

Let’s do one more example together before we go.

\(-3x+20=2\)

 

So, first we’re going to undo this addition by subtracting 20 from both sides.

\(-3x+20-20=2-20\)

 

That leaves us with:

\(-3x=-18\)

 

And then we need to get rid of this -3 being multiplied by \(x\), so we divide both sides by -3.

\(\frac{-3x}{-3}=\frac{-18}{-3}\)

 

This leaves us with:

\(x=6\)

 

So x=6 is our answer. And there you have it!

I hope that this video helped you better understand how to solve two-step equations. Thanks for watching and happy studying!

Practice Questions

Question #1:

 
Solve the equation for \(x\): \(4x-5=19\).

\(x=14\)
\(x=18\)
\(x=6\)
\(x=20\)
Answer:

To solve the equation for x, we want to undo any operations around x in order to isolate x on the left side of the equation. When solving two-step equations, always undo the addition or subtraction operation first, then undo the multiplication or division operation.

Five is being subtracted from 4x. The opposite operation is addition. We need to add 5 to both sides of the equation.

\(4x-5+5=19+5\)
\(4x=24\)

The number next to x implies we are multiplying it by \(x\), so we need to divide \(4x\) by 4 to undo the multiplication operation. Also, to keep our equation balanced, we need to divide by 4 on the right side of the equation.

\(\frac{4x}{\mathbf{4}}=\frac{24}{\mathbf{4}}\)

\(x=6\)

Question #2:

 
Solve the equation for \(x\): \(\frac{x}{3}+7=12\).

\(x=15\)
\(x=36\)
\(x=5\)
\(x=57\)
Answer:

To solve the equation for \(x\), we want to undo any operations around \(x\) in order to isolate \(x\) on the left side of the equation. When solving two-step equations, always undo the addition or subtraction operation first, then undo the multiplication or division operation.

In the equation, 7 is being added to \(\frac{x}{3}\). The opposite operation is subtraction, so we need to subtract 7 from both sides of the equation.

\(\frac{x}{3}+7-7=12-7\)

\(\frac{x}{3}=5\)

The fraction bar on the left side of the equation means we are dividing \(x\) by 3, so we need to multiply \(\frac{x}{3}\) by 3 to undo the division operation. Also, to keep our equation balanced, we need to multiply 3 on the right side of the equation.

\(\mathbf{3\times}\frac{x}{3}=\mathbf{3\times}5\)

\(x=15\)

Question #3:

 
Solve the equation for \(x\): \(-6x+27=3\).

\(x=-24\)
\(x=-26\)
\(x=5\)
\(x=4\)
Answer:

To solve the equation for \(x\), we want to undo any operations around \(x\) in order to isolate \(x\) on the left side of the equation. When solving two-step equations, always undo the addition or subtraction operation first, then undo the multiplication or division operation.

In the equation, 27 is being added to \(-6x\). The opposite operation is subtraction, so we need to subtract 27 from both sides of the equation.

\(-6x+27-27=3-27\)
\(-6x=-24\)

The number next to x implies we are multiplying it by \(x\), so we need to divide \(-6x\) by –6 to undo the multiplication operation. Also, to keep our equation balanced, we need to divide by –6 on the right-hand side of the equation to solve for \(x\).

\(\frac{-6x}{\mathbf{-6}}=\frac{-24}{\mathbf{-6}}\)

\(x=4\)

Question #4:

 
You want to save money to buy a new coat that costs $100. You have a part-time job that pays you $8 an hour after all deductions are taken out. If you have already saved $28 to buy the coat, how many more hours do you need to work to save enough money to buy the coat?

8 hours

9 hours

12 hours

10 hours

Answer:

Let the variable \(x\) represent the number of hours you work. Multiplying $8 by the unknown number of hours you need to work gives you the amount of money you make from working. You have already saved $28, so we can write the expression \(8x+28\) to represent the total amount of money you will have saved after working \(x\) hours. You need to save $100 to buy the new coat, so we can write the equation below to find the number of hours you need to work.

\(8x+28=100\)

To solve the equation for \(x\), we want to undo any operations around \(x\) in order to isolate \(x\) on the left-hand side of the equation. When solving two-step equations, always undo the addition or subtraction operation first, then undo the multiplication or division operation second.

In the equation, 28 is being added to \(8x\). The opposite operation is subtraction, so we need to subtract 28 from both sides of the equation.

\(8x+28-28=100-28\)
\(8x=72\)

The number next to x implies we are multiplying it by x, so we need to divide 8x by 8 to undo the multiplication operation. Also, to keep our equation balanced, we need to divide by 8 on the right-hand side of the equation as well.

\(\frac{8x}{\mathbf{8}}=\frac{72}{\mathbf{8}}\)

\(x=9\)

Therefore, you need to work 9 hours to have enough money to buy the new coat.

Question #5:

 
You have counted ten less than one-half of the number of marbles that are in a bag. If you have counted 64 marbles, how many marbles are in the bag?

74 marbles

36 marbles

148 marbles

108 marbles

Answer:

Let the variable \(x\) represent the number of marbles in the bag. Subtracting 10 marbles from one-half of the number of marbles that are in the bag gives us the number of marbles that have been counted. The expression \(\frac{x}{2}-10\) represents this statement. Since you have counted 64 marbles, we can write the equation below to find the number of marbles that are in the bag.

\(\frac{x}{2}-10=64\)

To solve the equation for \(x\), we want to undo any operations around \(x\) in order to isolate \(x\) on the left side of the equation. When solving two-step equations, always undo the addition or subtraction operation first, then undo the multiplication or division operation second.

In the equation, 10 is being subtracted from \(\frac{x}{2}\). The opposite operation is addition, so we need to add 10 to both sides of the equation.

\(\frac{x}{2}-10+10=64+10\)

\(\frac{x}{2}=74\)

The fraction bar on the left side of the equation means we are dividing \(x\) by 2, so we need to multiply \(\frac{x}{2}\) by 2 to undo the division operation. To keep our equation balanced, we need to multiply 2 on the right side of the equation.

\(\mathbf{2\times}\frac{x}{2}=\mathbf{2\times}74\)

\(x=148\)

Therefore, there are 148 marbles in the bag.

 

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by Mometrix Test Preparation | This Page Last Updated: April 27, 2022