Solving Multi-Step Equations

Learning Objectives

Introduction

There are some equationsA mathematical statement that two expressions are equal. that you can solve in your head quickly. For example, what is the value of `y` in the equation `2y = 6`? Chances are you didn’t need to get out a pencil and paper to calculate that `y = 3`. You only needed to do one thing to get the answer, divide `6` by `2`.

Other equations are more complicated. Solving `4(1/3t+1/2)=6` without writing anything down is difficult! That’s because this equation contains not just a variableA letter or symbol used to represent a quantity that can change. but also fractions and terms A number or product of a number and variables raised to powers. `4x`, `-5y^2`, `6`, and `x^3y^4` are all examples of terms. inside parentheses. This is a multi-step equationAn equation that requires more than one step to solve., one that takes several steps to solve. Although multi-step equations take more time and more operations, they can still be simplified and solved by applying basic algebraic rules.

Using Properties of Equalities

Remember that you can think of an equation as a balance scale, with the goal being to rewrite the equation so that it is easier to solve but still balanced. The addition property of equality For all real numbers `a`, `b`, and `c`, if `a = b`, then `a + c = b + c`. If two expressions are equal to each other and you add the same value to both sides of the equation, the equation will remain equal. and the multiplication property of equalityFor all real numbers `a`, `b`, and `c`, `c` ≠ `0`: If `a = b`, then `ac = bc`. If two expressions are equal to each other and you multiply both sides of the equation by the same non-zero number, the equation will remain equal. explain how you can keep the scale, or the equation, balanced. Whenever you perform an operation to one side of the equation, if you perform the same exact operation to the other side, you’ll keep both sides of the equation equal.

If the equation is in the form, `ax + b = c`, where `x` is the variable, you can solve the equation as before. First “undo” the addition and subtraction, and then “undo” the multiplication and division.

Example

Problem

Solve `3y + 2 = 11`.

 

Answer

`y = 3`

 

 

Example

Problem

Solve `1/4x-2=3`.

 

 

Add `2` to both sides of the equation to get the term with the variable by itself, so `1/4 x=5`.

Multiply both sides of the equation by `4` to get a coefficient of `1` for the variable.

Answer

`x = 20`

 

If the equation is not in the form, `ax + b = c`, you will need to perform some additional steps to get the equation in that form.

In the example below, there are several sets of like terms Terms that contain the same variables raised to the same powers. For example, `3x` and `-8x` are like terms, as are `8xy^2` and `0.5xy^2`. . You must first combine all like terms.

Example

Problem

Solve `3x + 5x + 4 - x + 7 = 88`.

There are three like terms `3x`, `5x` and `-x` involving a variable.

 

Combine these like terms.

`4` and `7` are also like terms and can be added.

 

The equation is now in the form `ax + b = c.` So, we can solve `7x+11=88` as before.

Subtract `11` from both sides.

Divide both sides by `7`.

Answer

`x = 11`

 

Some equations may have the variable on both sides of the equal sign. We need to “move” one of the variable terms in order to solve the equation.

Example

Problem

Solve `6x + 5 = 10 + 5x`. Check your solution.

This equation has `x` terms on both the left and the right. To solve an equation like this, you must first get the variables on the same side of the equal sign.

 

You can subtract `5x` on each side of the equal sign, which gives a new equation: `x + 5 = 10.` This is now a one-step equation!

 

Subtract `5` from both sides and you get `x=5`.

Check

`6x+5=10+5x`

`6(5)+5=10+5(5)`

`30+5=10+25`

`35=35`

Check your solution by substituting `5` for `x` in the original equation.

This is a true statement, so the solution is correct.

Answer

`x = 5`

 

Here are some steps to follow when you solve multi-step equations.

Solving multi-step equations

 

  1. If necessary, simplify the expressions on each side of the equation, including combining like terms.
  2. Get all variable terms on one side and all numbers on the other side using the addition property of equality. `(ax + b = c` or `c = ax + b)`
  3. Isolate the variable term using the inverse operation or additive inverse (opposite) using the addition property of equality.
  4. Isolate the variable using the inverse operation or multiplicative inverse (reciprocal) using the multiplication property of equality to write the variable with a coefficient of `1`.
  5. Check your solution by substituting the value of the variable in the original equation.

 

The examples below illustrate this sequence of steps.

Example

Problem

Solve for `y`.

-`20y + 15 = 2-16y+11` 

``

 

 

 

 

 

 

 

 

Step `1`. On the right side, combine like terms: `2 + 11 = 13.`

 

Step `2`. Add `20y` to both sides to remove the variable term from the left side of the equation.

Step `3`. Subtract `13` from both sides.

Step `4`. Divide `4y` by `4` to solve for `y.`````````

Check

`-20y+15` `text()=text()` `2-16y+11`
`-20(1/2)+15` `=` `2-16(1/2)+11`
`-10+15` `=` `2-8+11`
`5` `=` `5`

 

 

Step `5`. To check your answer, substitute `1/2` for `y` in the original equation. The statement `5 = 5` is true, so `y=1/2`  is the solution.

Answer

`y=1/2`

 
       

Advanced Example

Problem

Solve `3y + 10.5 = 6.5 + 2.5y`. Check your solution.

`3y+10.5=6.5+2.5y`

 

This equation has `y` terms on both the left and the right. To solve an equation like this, you must first get the variables on the same side of the equal sign.

 

`3y` `text()+text()10.5` `text()=text()` `6.5` `text()+text()2.5y`
`text()-text()2.5y`       `text()-text()2.5y`
`0.5y` `text()+text()10.5` `=` `6.5`  

 

 

Add `-2.5y` to both sides so that the variable remains on one side only.

 

`0.5y` `text()+text()10.5` `text()=text()` `6.5`
  `text()-text()10.5`   `text()-text()10.5`
`0.5y`   `=` `text()-text()4`

 

 

Now isolate the variable by subtracting `10.5` from both sides.

 

`10(0.5y)` `text()=text()` `10(-4)`
`5y` `=` `text()-text()40`
`(5y)/5` `=` `(-40)/5`
`y` `=` `text()-text()8`

 

Multiply both sides by `10` so that `0.5y` becomes `5y`, then divide by `5.`

Check

 

`3y+10.5` `text()=text()` `6.5+2.5y`
`3(-8)+10.5` `=` `6.5+(-20)`
`-24+10.5` `=` `6.5+(-20)`
`text()-text()13.5` `=` `text()-text()13.5`

 

 

 

 

Check your solution by substituting `-8` in for `y` in the original equation.

This is a true statement, so the solution is correct.

Answer

`y = -8`

 
       

Advanced Question

Identify the step that will not lead to a correct solution to the problem.

`3a-11/2=-(3a)/2+25/2`

 

A) Multiply both sides of the equation by `2`.

 

B) Add `11/2` to both sides of the equation.

 

C) Add `(3a)/2` to the left side, and add `-3a` to the right side.

 

D) Rewrite `3a` as `(6a)/2`.

 

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Solving Equations Involving Parentheses, Fractions, and Decimals

More complex multi-step equations may involve additional symbols such as parentheses. The steps above can still be used. If there are parentheses, you use the distributive property of multiplication as part of Step `1` to simplify the expression. Then you solve as before.

The Distributive Property of Multiplication

 

For all real numbers `a`, `b`, and `c`, `a(b + c) = ab+ac.`

 

What this means is that when a number multiplies an expression inside parentheses, you can distribute the multiplication to each term of the expression individually. Then, you can follow the routine steps described above to isolate the variableA method for solving an equation that involves rewriting an equivalent equation in which the variable is on one side of the equation and everything else is on the other side of the equation. to solve the equation.

Example

Problem

Solve for `a`.

`4(2a + 3)=-3(a-1)+31`
`4(2a+3)` `=` `-3(a-1)+31`
`8a+12` `text()=text()` `-3a+3+31`


 

 

 

`8a` `text()+text()12` `text()=text()` `text()-text()3a` `text()+text()34`
`text()+text()3a`     `text()+text()3a`  
`11a` `text()+text()12` `=`   `text()+text()34`
  `text()-text()12`     `text()-text()12`
`(11a)/11`   `=`   `22/11`
  `a` `=` `2`  

 

Apply the distributive property to expand `4(2a + 3)` to `8a + 12` and `-3(a - 1)` to `-3a + 3`.

 

 

 

Combine like terms.

Add `3a` to both sides to move the variable terms to one side.

Subtract `12` to isolate the variable term.

Divide both terms by `11` to get a coefficient of `1`.

 

Answer

`a = 2`

 
       

In which of the following equations is the distributive property properly applied to the equation `2(y+3)=7`?

 

A) `y + 6 = 7`

 

B) `2y + 6 = 14`

 

C) `2y + 6 = 7`

 

D) `2y + 3 = 7`

 

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If you prefer not working with fractions, you can use the multiplication property of equality to multiply both sides of the equation by a common denominator of all of the fractions in the equation. See the example below.

Example

Problem

Solve `1/2x-3=2-3/4x` by clearing the fractions in the equation first.

``

`1/2x-3` `text()=text()` `2-3/4x`
`4(1/2x-3)` `=` `4(2-3/4x)`
`4(1/2x)-4(3)` `=` `4(2)-4(3/4x)`
`4/2x-12` `=` `8-12/4x`

 

 

 

 

`2x` `text()-text()12` `text()=text()` `8` `text()-text()3x`
`text()+text()3x`       `text()+text()3x`
`5x` `text()-text()12` `=` `8`  
  `text()+text()12`   `text()+text()12`  
`(5x)/5`   `=` `20/5`  
  `x` `=` `4`  

Multiply both sides of the equation by `4,` the common denominator of the fractional coefficients.

Use the distributive property to expand the expressions on both sides.

Multiply.

 

 

 

 

 

 

Add `3x` to both sides to move the variable terms to only one side.

Add `12` to both sides to move the constantA symbol that represents a quantity that cannot change. It can be a number, letter or a symbol. terms to the other side.

Divide to isolate the variable.

Answer

`x=4`

 

Of course, if you like to work with fractions, you can just apply your knowledge of operations with fractions and solve.

Example

Problem

Solve `1/2x-3=2-3/4x.`

 
`1/2x` `text()-text()3` `text()=text()` `2` `text()-text()3/4x`
`text()+text()3/4x`   `=`   `text()+text()3/4x`
`5/4x` `text()-text()3` `=` `2`  
  `text()+text()3`   `text()+text()3`  
`5/4x`   `=` `5`  

 

 

 

 

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`4/5*5/4x` `text()=text()` `4/5*5`
`x` `=` `20/5`
`x` `=` `4`

Add `3/4x` to both sides to get the variable terms on one side.

`1/2+3/4=2/4+3/4=5/4-3/4+3/4=0`

Add `3` to both sides to get the constant terms on the other side.   

 

 

 

 

 

To get a coefficient of `1`, multiply the variable term by its multiplicative inverse.

Answer

`x=4`

 
         

Advanced Example

Problem

Solve `1/2(2+a)=(3a+4)/3^2`. Check your solution.

`1/2(2+a)=(3a+4)/3^2`

Solving this equation will require multiple steps. Begin by evaluating `3^2 = 9.`

`1/2*2+1/2*a=(3a+4)/9`

`1+1/2a=(3a+4)/9`

`1+a/2=(3a+4)/9`

 

Now distribute the `1/2` on the left side of the equation.

 

`18(1+a/2)` `text()=text()` `18((3a+4)/9)`
`18*1+18*a/2` `=` `(18(3a+4))/9`
`18+(18a)/2` `=` `(18(3a+4))/9`
`18+(18a)/2` `=` `(9*2(3a+4))/9`
`18+9a*2/2` `=` `2(3a+4)*9/9`
`18+9a` `=` `2(3a+4)`
`18+9a` `=` `2*3a+2*4`
`18+9a` `=` `6a+8`

 

 

 

 

 

 

Multiply both sides of the equation by `18`, the common denominator of the fractions in the problem.

Use the distributive property to expand the expression on the left side.

Then remove a factor of `1` from both sides. On the left, you can think of `(18a)/2`  as `2/2*9a.` On the right, you can think of `(18(3a+4))/9` as `9/9*(2(3a+4))/1.`

Continue solving for `a` using the distributive property.

 

 

`18` `text()+text()9a` `text()=text()` `6a` `text()+text()8`
  `text()-text()6a`   `text()-text()6a`  
`18` `text()+text()3a` `=`   `8`
`text()-text()18`       `text()-text()18`
  `3a` `=`   `text()-text()10`
  `(3a)/3` `=`   `(-10)/3`
  `a` `=`   `-10/3`

 

 

Then isolate the variable, and solve the remaining one-step problem.

 

Check

`1/2[2+(-10/3)]` `text()=text()` `(3(-10/3)+4)/3^2`
`1/2[2+(-10/3)]` `=` `(-10+4)/9`
`1/2[6/3+(-10/3)]` `=` `(-10+4)/9`
`1/2(-4/3)` `=` `(-10+4)/9`
`(-4)/6` `=` `(-10+4)/9`
`(-4)/6` `=` `(-6)/9`
`-2/3*2/2` `=` `-2/3*3/3`
`-2/3` `=` `-2/3`


 

 

Check your solution by substituting `-10/3` in for `a` in the original equation.

 

 

 

 

 

 

 

 

 

 

 

 

This is a true statement, so the solution is correct.

Answer

`a=-10/3`

 

 

To clear the fractions from `1/3-(2y)/9=19`, we can multiply both sides of the equation by which of the following numbers?

 

`3text( )6text( )9text( )27`

 

A) `9`

 

B) `9` or `27`

 

C) `6`

 

D) `3` or `9`

 

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Regardless of which method you use to solve equations containing variables, you will get the same answer. You can choose the method you find easier! Remember to check your answer by substituting your solution into the original equation.

Just as you can clear fractions from an equation, you can clear decimals from the equation in the same way. Find a common denominator and use the multiplication property of equality to multiply both sides of the equation.

Example

Problem

Solve `0.4x - 0.25 = 1.75` by clearing the decimals first.

`0.4x-0.25=1.75`

`100(0.4x-0.25)=100(1.75)`

 

 

`40x` `text()-text()25` `text()=text()` `175`
  `text()+text()25`   `text()+text()25`
`(40x)/40`   `=` `200/40`
  `x` `=` `5`

 

 

 

 

 

 

`0.4(4/10)` and `0.25(25/100)`  and `1.75(175/100)`have a common denominator of `100.`

 

Multiply both sides by `100.`

 

 

 

Apply the distributive property to clear the parentheses.

Solve as before. Add `25` to both sides.

Divide both sides by `40.`

 

 

Check:

`0.4x-0.25=1.75`

`0.4(5)-0.25=1.75`

`2-1.25=1.75`

`1.75=1.75`

 

Substitute `x = 5` into the original equation.

 

Evaluate.

The solution checks.

 

Answer

`x=5`

 

 

Advanced Question

Solve for `a`: `1/4(a+3)=2-a`

 

A) `a = 2`

 

B) `a = 1`

 

C) `a = 0`

 

D) `a = -2`

 

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Summary

Complex, multi-step equations often require multi-step solutions. Before you can begin to isolate a variable, you may need to simplify the equation first. This may mean using the distributive property to remove parentheses, or multiplying both sides of an equation by a common denominator to get rid of fractions. Sometimes it requires both techniques.