Equations

Ontario 10 Applied (MFM2P)

Rearranging then solving III

Lesson

Just as in Separating Letters & Numbers, the way to solve these equations is by combining the like terms so we have all the variables on one side and all the numbers on the other.

There is more than one way to solve these problems 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 equations remain equivalent.

Before considering the equations in the examples below, recall how you would solve a simpler equation such as $x-5=10$`x`−5=10.

We need to remove -5 from the left hand side to make $x$`x` the subject. How do we do that? By REVERSING the operation. The opposite of subtracting 5 would be adding 5, and by doing that we can cancel it out.

$x-5+5=10+5$`x`−5+5=10+5

$x=15$`x`=15

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

**Solve**: $5x+9=x+21$5`x`+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 numbers on the right hand side. That is, we want to move 9 to 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

Note: How can we test our solution?

**Solve**: $116-6x=12x+8$116−6`x`=12`x`+8

**Think:** Combine like terms. It doesn't matter which side you decide to place the variables on.

First, move variables to one side. In this case, let's move them to the right hand side so the coefficient of $x$`x` will be positive. We do this by adding $6x$6`x` to both sides.

$116-6x+6x$116−6x+6x |
$=$= | $12x+8+6x$12x+8+6x |

$116$116 | $=$= | $18x+8$18x+8 |

Now, group the constant numbers on the left hand side, by subtracting 8 from both sides.

$116-8$116−8 | $=$= | $18x+8-8$18x+8−8 |

$108$108 | $=$= | $18x$18x |

To solve for $x$`x`, we divide both sides by 18.

$x=6$`x`=6

**Solve:** $-28-6x=-12x+8$−28−6`x`=−12`x`+8

**Think:** We need to group the like terms so we have all the variables on one side and all the numbers on the other.

Let's start by grouping the variables on the left hand side of the equation. That is, move $-12x$−12`x` to the left hand side.

Then we move the numbers to the right hand side.

$-28-6x$−28−6x |
$=$= | $-12x+8$−12x+8 |

$-28-6x+12x$−28−6x+12x |
$=$= | $-12x+8+12x$−12x+8+12x |

$-28+6x$−28+6x |
$=$= | $8$8 |

$-28+6x+28$−28+6x+28 |
$=$= | $8+28$8+28 |

$6x$6x |
$=$= | $36$36 |

$x$x |
$=$= | $36$36 |

As we know, there is more than one way to solve a problem. What happens if we were to move the variables to the right hand side this time?

**Solve:** $-28-6x=-12x+8$−28−6`x`=−12`x`+8

To move the variables to the right hand side, we add $6x$6`x` to both sides. Then we move the numbers to the left hand side.

$-28-6x+6x$−28−6x+6x |
$=$= | $-12x+8+6x$−12x+8+6x |

$-28$−28 | $=$= | $-6x+8$−6x+8 |

$-28-8$−28−8 | $=$= | $-6x+8-8$−6x+8−8 |

$-36$−36 | $=$= | $-6x$−6x |

$x$x |
$=$= | $6$6 |

So you can see our answer is the same as in Question 3, even though we solved the equation slightly differently.

Solve the following equation for $r$`r`: $10r+8=74-r$10`r`+8=74−`r`

Solve for the value of $x$`x`: $4\left(5x+1\right)=-3\left(5x-5\right)$4(5`x`+1)=−3(5`x`−5)

Solve first-degree equations involving one variable, including equations with fractional coefficients