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# Solving equations using inverses (mixed operations)

## 9.3 Solving equations using inverses (mixed operations)

Often you need to simplify an equation before trying to solve it. To do this, collect the like terms and do some calculations.

## Worked example 9.8: Simplifying linear equations

Solve for $$r$$:

$- 4r + 2r + 9 = - 4 + 3$

### Collect the like terms.

We need to simplify each side of the equation before trying to solve the equation. We can look at each side of the equation separately to see if we can add any like terms.

\begin{align} - 4r + 2r + 9 &= - 4 + 3 \\ \therefore - 2r + 9 &= - 1 \end{align}

### Solve the equation.

Now the equation should look familiar, and we are going to do the same thing to both sides of the equation to get $$r$$ on its own.

We want to get rid of the $$9$$ that is on the same side as $$r$$. To do this we must subtract $$9$$ from both sides.

\begin{align} - 2r + 9 - 9 &= - 1 - 9 \\ \therefore - 2r &= - 10 \end{align}

The last step is to divide by the coefficient of $$r$$ on both sides:

\begin{align} \therefore - \frac{2r}{- 2} &= - \frac{10}{- 2} \\ \therefore r &= 5 \end{align}

## Worked example 9.10: Simplifying linear equations with variables on both sides

Solve for $$r$$:

$- 4r - 3 = - 6r + 4$

### Get the $$r$$’s on the left-hand side.

We need to do the same thing to both sides to get all of the $$r$$’s onto the left-hand side of the equation. Then the equation will look like the equations that we can already solve.

We want to get rid of $$- 6r$$ on the right-hand side so that we only have $$r$$’s on the left-hand side of the equation. To do so, we must add $$6r$$ to both sides.

\begin{align} - 4r - 3 &= - 6r + 4 \\ 4r - 3 + 6r &= - 6r + 4 + 6r \\ 2r - 3 &= 4 \end{align}

### Solve the equation using inverse operations.

The equation should now look familiar. We can solve it by doing the same thing to both sides, to get $$r$$ on its own.

\begin{align} 2r - 3 &= 4 \\ 2r - 3 + 3 &= 4 + 3 \\ 2r &= 7 \\ \frac{2r}{2} &= \frac{7}{2} \\ \therefore r &= \frac{7}{2} \end{align}

## Worked example 9.11: Simplifying linear equations (order of operations)

Solve for $$r$$:

$- 16 = - 6 - 2(r + 2)$

### Simplify the right-hand side.

To simplify any expression, we must first separate it into terms. Then, we simplify each term. Finally, we collect like terms.

Terms are separated by + and − and joined by × and ÷. Separating expressions into terms will make sure that we do the operations in the correct order.

$- 16 = - 6 - 2(r + 2)$

There are two terms on the right hand side of the equation: $$-6$$ and $$-2(r + 2)$$. We must first simplify $$- 2(r + 2)$$ by distributing in the $$-2$$:

\begin{align} - 16 &= - 6 - 2r - 4 \\ - 16 &= - 2r - 10 \end{align}

### Solve the equation with inverse operations.

\begin{align} - 16 &= - 2r - 10 \\ - 16 + 10 &= - 2r - 10 + 10 \\ - 6 &= - 2r \\ \frac{-6}{-2} &=\frac{-2r}{-2} \\ \therefore 3 &= r \\ \therefore r &= 3 \end{align}

There are many different ways that you can solve linear equations. But these three steps will always get you to the correct answer:

• Simplify each side of the equation (remember the order of operations).
• Get the variables on the left-hand side.
• Use inverse operations to isolate the variable.

## Worked example 9.12: Simplifying linear equations (order of operations)

Solve for $$y$$:

$- 7 + 5(y + 3) = 7y - 2(2 + 4y)$

### Simplify each side of the equation.

First we must simplify each side, if possible. In this question, we can use the distributive law to multiply out the brackets, and then collect like terms.

\begin{align} - 7 + 5(y + 3) &= 7y - 2(2 + 4y) \\ - 7 + 5y + 15 &= 7y - 4 - 8y \\ 8 + 5y &= - y - 4 \end{align}

### Get the variables onto the left-hand side of the equation.

Now that we have simplified each side of the equation, it looks much more manageable. The next step is to get the $$y$$’s on the left hand side of the equation. We want to eliminate the $$-y$$ from the right-hand side, so we need to add $$y$$ to both sides of the equation.

\begin{align} 8 + 5y &= - y - 4 \\ 8 + 5y + y &= - y - 4 + y \\ 8 + 6y &= - 4 \end{align}

### Use inverse operations to isolate $$y$$.

\begin{align} 8 + 6y &= - 4 \\ 8 + 6y - 8 &= - 4 - 8 \\ 6y &= - 12 \\ \frac{6y}{6} &= - \frac{12}{6} \\ \therefore y &= - 2 \end{align}
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