An equation shows that two expressions are equal in value. Because of this equivalence we can use a balance scale model to visualize how to solve an equation.

Consider the balance scale model for the equation 2(2x-1)=2(x+5) where a blue bag represents x, a sack of balloons tied to the scale represents -x and takes away the same amount of weight that a blue bag adds. A blue block represents +1 and a single balloon tied to the scale represents -1 and takes away the same amount of weight that a blue block adds.

Since each side of the balance scale is made up of two equal groups of something, we can eliminate one group from each side. Remember, the scale is balanced so the weight on each side was equal and now we have removed half of the weight from each side so the scale will stay balanced.

This gives us an equation of 2x-1=x+5 which is equivalent to the original equation.

On the left side of the balanced equation, we have a balloon (negative). We can create a zero pair by adding a blue (positive) block. Adding a positive block to the left side will increase the weight of that side and throw off the balance. So we need to also add a blue (positive) block to the right side to keep the scale balanced.

This creates an equation of 2x-1+1=x+5+1 which is equivalent to both of the previous equations.

We can remove the zero pair (one blue block and one balloon) from the left side of the scale without changing the weight and get an equation of 2x=x+6.

On the right side of the balanced equation, we have a blue (positive) bag. We can create a zero pair by adding a bag of balloons to the right side of the scale and to keep the scale balanced will add a bag of balloons to the left side as well.

This gives us an equation of 2x-x=x-x+6. This new equation is equivalent to all previous equations.

Removing the zero pairs on the left and right sides of the scale, we get a new equation x=6. This means that one blue bag is equal to six blue (positive) blocks.

We have now solved the equation 2(2x-1)=2(x+5) and got x=6 using a balance scale.

Consider the equation 4(xâˆ’3)+8=12

a

Represent the equation using algebra tiles.

Worked Solution

b

Use the algebra tiles to solve the equation.

Worked Solution

A local community library is organizing a book drive. Jordan collects 5 boxes of books and 2 individual books. A local bookstore donates 3 boxes of books and 12 individual books to the drive.

Each box contains an equal number of books, and Jordan and the bookstore both donated the same total number of books.

a

Write an equation to represent this situation.

Worked Solution

b

Represent the equation using a pictorial model.

Worked Solution

c

Use the pictorial model to solve the equation.

Worked Solution

Idea summary

We can use visual models to represent and solve equations pictorally. We can use **zero pairs** and applying the same changes to both sides to maintain the balance or equality of the equation.

While visual models help us to better understand equations and their solutions, it's not always easy or efficient to solve equations using visual models.

Fortunately, we can use **inverse operations**, the **properties of real numbers**, and the **properties of equality** to solve equations algebraically.

Recall the properties of equality, inverse, and identity:

Symmetric property of equality | \text{If } a=b, \text{then } b=a |
---|---|

Transitive property of equality | \text{If } a=b \text{ and } b=c, \text{then } a=c |

Addition property of equality | \text{If } a=b, \text{then } a+c=b+c |

Subtraction property of equality | \text{If } a=b, \text{then } a-c=b-c |

Multiplication property of equality | \text{If } a=b, \text{then } ac=bc |

Division property of equality | \text{If } a=b \text{ and } c \neq0, \text{then } \dfrac{a}{c}=\dfrac{b}{c} |

Substitution property of equality | \text{If } {a=b}, \text{ then } b \text{ may be substituted for } a \text{ in any expression} |

Additive identity | \text{If } {a=b}, \text{ then } a+0=b \text{ and } a=b+0 |

Multiplicative identity | \text{If } {a=b}, \text{ then } a\cdot 1=b \text{ and } a=b \cdot 1 |

Additive inverse | a+(-a)=0 |

Multiplicative inverse | a \cdot \dfrac{1}{a}=1 |

When we solve equations with variables on both sides, we want to rearrange the equation so that all of the variables are on the same side of the equation and all of the constants are on the other side of the equation. This process is similar to combining like terms, we just do it across the equal sign.

The side you choose to move variables to does not make a difference to the final solution; however, there is usually a way that will make it easier to solve.

For example, if we were solving the equation:

\displaystyle 10x - 4 | \displaystyle = | \displaystyle 6x | Given equation |

\displaystyle 10x-6x-4 | \displaystyle = | \displaystyle 6x-6x | Subtraction property of equality |

\displaystyle 4x-4 | \displaystyle = | \displaystyle 6x-6x | Combine like terms |

\displaystyle 4x-4 | \displaystyle = | \displaystyle 0 | Additive inverse |

\displaystyle 4x-4+4 | \displaystyle = | \displaystyle 0+4 | Addition property of equality |

\displaystyle 4x | \displaystyle = | \displaystyle 0+4 | Additive inverse |

\displaystyle 4x | \displaystyle = | \displaystyle 4 | Additive identity |

\displaystyle \dfrac{4x}{4} | \displaystyle = | \displaystyle \dfrac{4}{4} | Division property of equality |

\displaystyle 1 \cdot x | \displaystyle = | \displaystyle 1 | Multiplicative inverse |

\displaystyle x | \displaystyle = | \displaystyle 1 | Multiplicative identity |

Solve the following equations:

a

5 \left( 2 y + 2\right) + 3 \left( 4 y - 5\right) +5y= 45

Worked Solution

b

- \dfrac{3}{2} \left(x + \dfrac{5}{4}\right) + \dfrac{11}{3} = - \dfrac{125}{6}

Worked Solution

c

9x+40-7x=4x

Worked Solution

d

12.6x+28.25=1.1x+85.75

Worked Solution

e

2\left(x-3\right)=x-3.

Worked Solution

Idea summary

When solving multi-step equations involving parentheses:

- Use the distributive property to clear parentheses.
- Combine like terms to simplify each side.
- Add or subtract to get the variable term on one side of the equals sign.
- Multiply or divide to isolate the variable and solve the equation.

To solve an equation with variables on both sides, we can rearrange the equation so that all of the variables are on the same side of the equation and all of the constants are on the other side of the equation.