United States of AmericaPA
High School Core Standards - Algebra I Assessment Anchors

# 3.03 Solving multi-step equations

Lesson

## Solving multi-step equations

When we want to solve linear equations, we are basically trying to find an unknown value.

Regardless of the operations involved in an equation, our aim is always the same. That is, to find the value of the unknown by 'getting it on its own'. We can do this in a variety of ways by applying the properties of equality.

#### Worked examples

##### Question 1

Solve the following equation: $-x-\frac{7}{8}=3$x78=3.

Think: We need to get $x$x by itself, which we will do in a couple of steps.

Do:

 $-x-\frac{7}{8}$−x−78​ $=$= $3$3 (multiply each term by $8$8 to get rid of the denominator) $-8x-7$−8x−7 $=$= $24$24 (add $7$7 to both sides) $-8x$−8x $=$= $31$31 (divide both sides by $-8$−8) $x$x $=$= $\frac{-31}{8}$−318​

Reflect:  What properties of equality were applied in each step?

##### Question 2

Solve the following equation: $-7\left(x+1\right)=-28$7(x+1)=28.

Think: Again we aim to get $x$x by itself, which we will do in a couple of steps.

Do:

 $-7\left(x+1\right)$−7(x+1) $=$= $-28$−28 (divide both sides by $-7$−7) $x+1$x+1 $=$= $\frac{-28}{-7}$−28−7​ $x+1$x+1 $=$= $4$4 (subtract $1$1 from both sides) $x$x $=$= $3$3

Reflect:  How might you justify your solution path using the properties of equality?

#### Practice questions

##### Question 3

Solve the following equation for $v$v:

$12v+6=110-v$12v+6=110v

##### Question 4

Solve the following equation: $5x-\frac{104}{5}=x$5x1045=x

## Solve equations with variables on both sides

There is more than one way to solve an equation, but it's important to remember that any operation must be applied to BOTH sides of an equation. What you do to one side, you must do to the other so that the values remain equivalent.

In solving each equation below, the priority will be to group like terms. To do this, ask yourself what operation you must reverse.

#### Worked examples

##### Question 5

Solve: $5x+9=x+21$5x+9=x+21

Think and Do: We have two choices in combining like terms: We could move the variables to the left and the constants to the right, OR we could move the variables to the right and the constants to the left. It would be nice to be left with a positive coefficient of $x$x

Let's have the variables on the left-hand side, that is, let's move $x$x to the left-hand side. To do this, we subtract $x$x from both sides.

 $5x+9-x$5x+9−x $=$= $x+21-x$x+21−x $4x+9$4x+9 $=$= $21$21

Now, we want the constant terms on the right-hand side. That is, we the term $9$9 to be on the right-hand side. To do this, we subtract 9 from both sides.

 $4x+9-9$4x+9−9 $=$= $21-9$21−9 $4x$4x $=$= $12$12

We can now solve for $x$x by dividing both sides by 4 to give us the solution.

$x=3$x=3

Reflect: Can you think of another way that we can solve the same equation in different steps?  Test that your alternate path yields the same solution.

#### Practice questions

##### Question 6

Solve the following equation for $r$r:

$8r-2=4r+14$8r2=4r+14

##### Question 7

Solve: $4\left(5x+1\right)=-3\left(5x-5\right)$4(5x+1)=3(5x5)

## How many solutions are there?

We have learned how to determine whether or not a value satisfies an equation. To determine this, we need to check whether the left-hand side (LHS) of the equation is equal to the right-hand side (RHS) of the equation. However, sometimes an equation may not just have one answer- sometimes there may be an infinite number of solutions and sometimes there may be no solutions.

For example, what happens if we had an equation $10=-4$10=4? We know this is not a solution. In fact, this is an example of an equation that has no solutions.

What happens if we have an equation $x=x$x=x? If we substituted values into this equation, we could say $1=1$1=1, $2=2$2=2, $3=3$3=3 and so on. This is an example of an equation that has an infinite number of solutions.

If we get a solution to an equation such as $x=5$x=5, this means that $5$5 is the only value that satisfies the equation. It is example of an equation with one solution.

Summary
• An equation of the form $x=a$x=a, where $x$x is a variable and $a$a is a number, has one solution.
• An equation of the form $a=a$a=a has infinitely many solutions.
• An equation of the form $a=b$a=b, where $a$a and $b$b are numbers, has no solutions.

#### Worked examples

##### Question 8

Solve: How many solutions does $\frac{10+x}{10}=\frac{x+10}{10}$10+x10=x+1010 have?

Think: How can we transfer the given equation into a simpler form ($x=a$x=a,$a=a$a=a or $a=b$a=b)?

Do:

 $\frac{10+x}{10}$10+x10​ $=$= $\frac{x+10}{10}$x+1010​ Multiply both sides by $10$10 to remove the denominators $\frac{10\left(10+x\right)}{10}$10(10+x)10​ $=$= $\frac{10\left(x+10\right)}{10}$10(x+10)10​ Simplify $10+x$10+x $=$= $x+10$x+10

Reflect: Can you see that the LHS of this equation is the same as the RHS? They are equivalent expressions but the terms are just switched around. This means this equation has an infinite number of solutions.

##### QUESTION 9

Solve: How many solutions does $x=x-10$x=x10 have?

Think: Can a number ever equal the same number minus $10$10?

Do: This equation is in the form $a=b$a=b because a number can never equal the same number minus $10$10. Therefore, there are no solutions to this equation.

Another way to see that there are no solutions is to complete the equation solving process.

 $x$x $=$= $x-10$x−10 $x-x$x−x $=$= $x-10-x$x−10−x subtract x from both sides $0$0 $=$= $-10$−10 because $0$0 is not equal to $-10$−10, we now know there are no solutions.

Reflect:  Can you write another equation that has no solutions?

#### Practice questions

##### Question 8

How many solutions does $18x=18x$18x=18x have?

1. infinitely many

A

no solutions

B

one solution

C

infinitely many

A

no solutions

B

one solution

C

##### question 9

How many solutions does the equation $10\left(5+x\right)=10\left(x+5\right)$10(5+x)=10(x+5) have?

1. Infinitely many

A

No solutions

B

One solution

C

Infinitely many

A

No solutions

B

One solution

C

### Outcomes

#### CC.2.2.HS.D.9

Use reasoning to solve equations and justify the solution method.

#### A1.1.2.1.1

Write, solve, and/or apply a linear equation (including problem situations).

#### A1.1.2.1.2

Use and/or identify an algebraic property to justify any step in an equation-solving process. Note: Linear equations only.