# Multi-step Equations

**Multi-step equations** are algebraic expressions that require more than one operation, such as subtraction, addition, multiplication, division, or exponentiation, to solve. It is important to know about the order of operations when solving multi-step equations.

Solve \(2x + 4 = 10\) for \(x\).

In order to solve for \(x\), we must isolate \(x\). To do this, we need to get rid of the plus \(4\) and the \(2\) in \(2x\) in such a way that the equation remains true (both sides of the equation are still equal).To do this, we first subtract \(4\) from both sides of the equation:

\[ \begin{align} 2x+4 &= 10 \\ 2x &= 6. \end{align}\]

Then, to isolate \(x\), we need to divide both sides by \(2:\)

\[ \begin{align} 2x &= 6 \\ x &= 3. \end{align}\]

Since multiple steps are required to isolate \(x\), this is a multi-step equation.

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## Methods for Solving Multi-step Equations

Solving multi-step equations uses combinations of addition, subtraction, multiplication, and division to isolate the variable in question.

Inverse OperationsIn order to solve equations, it is often necessary to isolate the variable in question. To do this, cancel out operations on the variable of interest while maintaining the equality of the equation. This can be done by performing inverse operations on the terms that need to be removed so that the variable of interest is isolated. Subtraction cancels out addition, and vice versa, and multiplication cancels out division, and vice versa.

Given an equation with a single unknown variable \(x\), solve for \(x\) by performing algebraic operations on both sides of the equation in order to isolate \(x\). The goal is to choose a sequence of algebraic operations that will leave \(x\) on one side and put everything else on the other side of the equal sign. Operations that can be used on both sides of the equation include the following:

- Adding or subtracting a constant

For example, \(x + 2 = 5 \implies x + 2 - 2 = 5 - 2 \implies x = 3.\)

- Multiplying or dividing by a constant

For example, \(2x = 4 \implies \frac{2x}{2} = \frac{4}{2} \implies x = 2.\)

- Adding or subtracting a multiple of \(x\) to collect terms

For example, \(3x + 4x = 5 \implies 7x = 5\) or \(2x + 4 = x + 5 \implies 2x - x + 4 = x + 5 - x \implies x + 4 = 5 \implies x + 4 - 4 = 5 - 4 \implies x = 1.\)

- Taking the root of an exponent or exponentiating a root

For example, \(x = y^2 \implies y = \pm \sqrt{x}.\)

To make computations easier, it may be a good idea to first separate the terms involving \(x\) from the constants by performing the steps of subtraction or addition before performing multiplication or division to remove the coefficient in front of the variable. It is important to pay careful attention to order of operations when solving multi-step equations.

Solve the following equation for \(x: 2y + 1 = 2x - 3 \).

We need to isolate \(x\), so we can start moving terms on the right side of the equals sign to the left side of the equals sign. To get rid of the minus \(3\) term of \(2x - 3\), we need to add \(3\). However, in order to maintain the equality of the expression, we need to add \(3\) tobothsides of the equation:\[2y + 1 + 3 = 2x - 3 + 3 = 2y + 4 = 2x.\]

Now we need to transform the equation so that we get rid of the \(2\) times part of \(2x\). To cancel out multiplication, use division. To do this, divide both sides of the equation by \(2:\)

\[\frac{2y+4}{2} = x.\]

We can simplify this further:

\[y + 2 = x,\]

which is equivalent to

\[x = y + 2.\]

We have successfully solved for \(x.\) \(_\square\)

## Multi-step Equations - Basic

What value of \(x\) satisfies

\[ 3x - 2 = x + 4?\]

We have\[ \begin{align} 3x - 2 &= x + 4 \\ 3x -2 + 2 &= x + 4 + 2 && \text{(adding 2 to both sides)} \\ 3x &= x + 6 \\ 3x - x &= x + 6 - x && \text{(subtracting } x \text{ from both sides)}\\ 2x &= 6 & \\ x &= 3. \qquad && \text{(dividing both sides by 2)} \end{align}\]

Therefore, \(x=3\). \( _\square \)

What value of \(x\) satisfies

\[ \frac{2}{3}x +6 = x ?\]

We have\[ \begin{align} \frac{2}{3}x +6 &= x \\ \frac{2}{3}x +6 - \frac{2}{3}x &= x - \frac{2}{3}x && \left(\text{subtracting } \frac{2}{3}x \text{ from both sides}\right) \\ 6 &= \frac{1}{3}x \\ 6 \cdot 3 &= \frac{1}{3}x \cdot 3 && (\text{multiplying both sides by 3})\\ 18 &= x . \end{align}\]

Therefore, \(x=18\). \( _\square \)

What value of \(x\) satisfies

\[ 3(2x-4)= 2(x+4) ?\]

We have\[ \begin{align} 3(2x-4) &= 2(x+4) \\ 6x - 12 &= 2x + 8\\ 6x - 2x - 12 &= 8 && (\text{subtracting } 2x \text{ from both sides}) \\ 4x - 12 &= 8\\ 4x &= 20 && (\text{adding 12 to both sides})\\ x &= 5. && (\text{dividing both sides by 4}) \end{align}\]

Therefore, \(x=5\). \( _\square \)

What number satisfies the property that when three times the number is increased by \(2,\) it has the same value as four times the number decreased by \(8?\)

Let \(x\) denote the number satisfying the property. Then three times the number increased by \(2\) is \(3x + 2\), and four times the number decreased by \(8\) is \(4x-8\). Since these values are equal, the equation we would like to solve is\[ \begin{align} 3x+2 &= 4x - 8 \\ 2 &= 4x - 8 - 3x\\ 2 &= x - 8\\ 2 + 8 &= x\\ 10 &= x. \end{align}\]

Therefore, \(x=10\). \( _\square \)

What value of \(x\) satisfies

\[ \frac{3x+5}{4}= \frac{x}{3} - \frac{x}{4}?\]

We have\[ \begin{align} \frac{3x+5}{4} &= \frac{x}{3} - \frac{x}{4} \\ \frac{3x+5}{4} \cdot 12 &= \frac{x}{3} \cdot 12 - \frac{x}{4} \cdot 12 \qquad (\text{multiplying both sides by 12})\\ (3x+5)(3) &= (x\cdot 4) - (x \cdot 3) \\ 9x + 15 &= 4x - 3x \\ 9x + 15 &= x \\ 8x & = -15 \\ x &= -\frac{15}{8}. \end{align}\]

Therefore, \(x= - \frac{15}{8}\). \( _\square \)

## Multi-step Equations - Intermediate

## See Also

**Cite as:**Multi-step Equations.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/multi-step-equations/