Solving Multi-Step Equations

Hello! Today we’re going to take a look at some examples of solving multi-step equations. Solving multi-step equations involves combining all our other equation-solving techniques. So let’s work through some problems so we can get more comfortable with solving equations. Let’s start with:

\(2(3x+4)-7=21x-9\)

 

We’re going to start by distributing this 2 on the left side. So, if we distribute the 2 to this set of parentheses:

\(6x+8-7=21x-9\)

 

Now we’re going to combine like terms on the left side of our equation. So, over here, our like terms are 8 and -7.

\(6x+1=21x-9\)

 

Now we want to get all of our variable terms on one side and all of our constant terms on the other. So to do that, we will start by subtracting \(6x\) from both sides.

\(6x-6x+1=21x-6x-9\)
 
\(1=15x-9\)

 

Now we want to get our constants on the left side, so we’ll do that by adding 9 to both sides. So this gives us:

\(1+9=15x-9+9\)
 
\(10=15x\)

 

And then to get \(x\) by itself we divide both sides by 15.

\(\frac{10}{15}=\frac{15x}{15}\)
 
\(\frac{10}{15}=x\)

 

Now, this fraction isn’t in simplest form, so we need to simplify that before it’s our final answer. We can simplify by dividing both the numerator and denominator by 5.

\(x=\frac{10}{15}=\frac{10\div 5}{15\div 5}=\frac{2}{3}\)

 

So we have \(x=23\).

Let’s take a look at another problem.

\(2x+10=4(6x+8)\)

 

For this problem, we’re going to start by distributing on the right side. So our left side is going to stay the same for right now.

\(2x+10=24x+32\)

 

Now I’m going to subtract \(2x\) from both sides. If you wanted to, you could subtract \(24x\) from both sides and get the \(x\)-terms on the left side. I’m going to get the \(x\)-terms on the right side for this problem.

\(2x-2x+10=24x-2x+32\)
 
\(10=22x+32\)

 

Now to get our constants on the left side, we’ll need to subtract 32 from both sides.

\(10-32=22x+32-32\)
 
\(-22=22x\)

 

And then to get \(x\) by itself, we divide both sides by 22.

\(\frac{-22}{22}=\frac{22x}{22}\)
 
\(-1=x\)

 

Before we go, let’s try one last problem together. This one’s going to look a lot more complicated, but we’ll still be able to solve it using all the same techniques we have been.

\(\frac{1}{2}(4x+16)-24+3x=2(x+3)-1\)

 

So again, we see we have distribution, this time on both sides of the equation, so we’re going to do that first.

\(2x+8-24+3x=2x+6-1\)

 

Now we’re going to combine like terms on both sides of the equation. So, on the left side, our like terms are 8 and -24, and \(2x\) and \(3x\). And then on the right side, our like terms are 6 and -1.

\(5x-16=2x+5\)

 

Now I’m going to get all of the variable terms on one side and all the constants on the other side. So, I’m going to start by adding 16 to both sides of the equation.

\(5x-16+16=2x+5+16\)
 
\(5x=2x+21\)

 

And then we’re going to get the \(x\)-terms on the left side, so subtract \(2x\) from both sides.

\(5x-2x=2x-2x+21\)
 
\(3x=21\)

 

And to solve for \(x\) and get \(x\) by itself, all we have to do is divide both sides by 3.

\(\frac{3x}{3}=\frac{21}{3}\)
 
\(x=7\)

 

And that’s all there is to it! I hope that working through some of these problems made things a little bit more clear for you. Thanks for watching and happy studying!

Practice Questions

Question #1:

 
Solve the equation for \(x\): \(4x+10=6(2x-5)\).

\(x=5\)
\(x=8\)
\(x=-4\)
\(x=-9\)
Answer:

Distribute 6 on the right side of the equation.

\(4x+10=12x-30\)

We can get the variable terms on the right side of the equation by subtracting \(4x\) from both sides.

\(4x+10-4x=12x-30-4x\)
\(10=8x-30\)

Now, add 30 to both sides of the equation.

\(10+30=8x-30+30\)
\(40=8x\)

Finally, divide both sides by 8.

\(\frac{40}{8}=\frac{8x}{8}\)
\(5=x\)

This is equivalent to \(x=5\).

Question #2:

 
Solve the equation for \(x\): \(3(2x+5)-6=18x+15\).

\(x=\frac{4}{3}\)
\(x=-\frac{1}{2}\)
\(x=\frac{2}{5}\)
\(x=-\frac{5}{3}\)
Answer:

Distribute 3 on the left side of the equation.

\(6x+15-6=18x+15\)

Then, combine like terms on the left side of the equation.

\(6x+9=18x+15\)

To get the variable terms on the left side, subtract \(18x\) from both sides of the equation.

\(6x+9-18x=18x+15-18x\)
\(-12x+9=15\)

Now, subtract 9 from both sides.

\(-12x+9-9=15-9\)
\(-12x=6\)

Then, divide both sides by –12.

\(\frac{-12x}{-12}=\frac{6}{-12}\)
\(x=-\frac{6}{12}\)

Reducing our answer by 6 gives the answer in simplest form.

\(x=-\frac{6\div6}{12\div6}=-\frac{1}{2}\)

Question #3:

 
Solve the equation for \(x\): \(\frac{1}{4}(12x+8)-16+5x=3(2x+4)-6\).

\(x=-8\)
\(x=-5\)
\(x=6\)
\(x=10\)
Answer:

Distribute \(\frac{1}{4}\) on the left side and 3 on the right side of the equation.

\(3x+2-16+5x=6x+12-6\)

Combine like terms on the left side and on the right side of the equation.

\(8x-14=6x+6\)

To get the variable terms on the left side, subtract \(6x\) from both sides of the equation.

\(8x-14-6x=6x+6-6x\)
\(2x-14=6\)

Now, add 14 to both sides.

\(2x-14+14=6+14\)
\(2x=20\)

Finally, divide both sides by 2.

\(\frac{2x}{2}=\frac{20}{2}\)
\(x=10\)

Question #4:

 
Five less than the product of 3 and the sum of a number \(x\) and 6 equals the sum of 24 and one-half the difference of the number and 2. What is the number \(x\)?

\(x=9\)
\(x=4\)
\(x=3\)
\(x=8\)
Answer:

Let \(x\) be the number you are trying to find. The product of 3 and the sum of \(x\) and 6 can be written by the algebraic expression \(3(x+6)\). Five less than this expression is \(3(x+6)-5\). One-half the difference of the number and 2 can be written by the algebraic expression \(\frac{1}{2}(x-2)\). The sum of 24 and this expression is \(24+\frac{1}{2}(x-2)\). Since the two algebraic expressions are the equal, we have the equation:

\(3(x+6)-5=24+\frac{1}{2}(x-2)\)

To find the number, distribute 3 on the left side and \(\frac{1}{2}\) on the right side of the equation first.

\(3x+18-5=24+\frac{1}{2}x-1\)

Combine like terms on the left and right side of the equation.

\(3x+13=23+\frac{1}{2}x\)

To get the variable terms on the left side, subtract \(\frac{1}{2}x\) from both sides of the equation.

\(3x+13-\frac{1}{2}x=23+\frac{1}{2}x-\frac{1}{2}x\)
\(\frac{5}{2}x+13=23\)

Now, subtract 13 from both sides.

\(\frac{5}{2}x+13-13=23-13\)
\(\frac{5}{2}x=10\)

Then, multiply both sides by \(\frac{2}{5}\).

\(\frac{2}{5}\cdot\frac{5}{2}x=10\cdot\frac{2}{5}\)
\(x=\frac{20}{5}\)
\(x=4\)

Question #5:

 
Suppose the revenue, in thousands of dollars, generated by a company that sells \(x\) units, where \(x\) is in hundreds of units, can be written by the algebraic expression \(4(x+1)-4\). The cost for producing the units can be written by the algebraic expression \(2(x-1)+8\). The break-even point occurs when the revenue generated by selling the units and cost to produce them are the same. How many units must be produced and sold for the company to reach the break-even point?

325

300

200

250

Answer:

To reach the break-even point, we need to set the revenue equal to the cost. This gives the equation:

\(4(x+1)-4=2(x-1)+8\)

To find the number of units to reach the break-even point, solve the equation. Distribute 4 on the left side and 2 on the right side of the equation.

\(4x+4-4=2x-2+8\)

Combine like terms on the left and right side of the equation.

\(4x=2x+6\)

To get the variable terms on the left side, subtract \(2x\) from both sides of the equation.

\(4x-2x=2x+6-2x\)
\(2x=6\)

Then, divide both sides by 2.

\(\frac{2x}{2}=\frac{6}{2}x=3\)

Since x represents the number of units produced in the hundreds, the company needs to produce \(3\times100=300\) units to break even.

 

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by Mometrix Test Preparation | This Page Last Updated: June 7, 2022