# Two-Step Equations Containing Integers

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Oftentimes, when we are confronted with a problem, we translate it first into an equation then use it to solve for the missing quantity. These equations may involve one, two or more steps before arriving at its solutions. Equations such as 2x + 3 = 5 and 3x - 5 = 7, require two–step solutions. These equations are called two–step equations.

These are equations that require two steps or two operations before arriving at its solutions. In solving two–step equations, there is no definite rule “which” operation to undo first. However, one can apply the rule for order of operations or whichever will result to a simpler equation. Again, the purpose in solving two-step equation is to combine the constants in one side of the equation and see to it that the variable must only have 1 as its numerical coefficient.

Before illustrating the steps on how to solve two-step equations, listed below are examples of two–step equations containing integers.

1. $2x-7=5$
2. $2n+ \left(-3\right ) =-1$
3. $5a-1=4$
4. $2a-\left(-2\right)=16$
5. $\displaystyle \frac{a}{3}+2=\left(-5\right)$

## Solving Two-Step Equations Containing Integers

To solve two–step equations, do the inverse of the operations involved one at a time. Since, we are dealing with equations, it must be noted that whatever operations are done on one side of the equation must as well be done on the other side. Also, we apply the rules for performing operations with integers. To illustrate, consider the following examples:

Solve for the unknown variable:

1. $2x-7=5$

Solution:

If we divide both sides by 2, we are going to have fractions. It is more preferable to undo subtraction first. Since twice the variable $x$ is subtracted by 7 and the opposite of addition is subtraction, then we add 7 to both sides of the equation. Then divide the resulting equation by 2.

2. $2x+(-3)=-1$

Solution:

Again, division of 2 in the equation will lead to fractions. We undo the addition first. Add the opposite of , which is 3 to both sides of the equation then divide both sides by 2.

$\begin{array}{c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c} &2x+\left ( -3 \right )&=&\left ( -1 \right ) \\ &2n+\left ( -3 \right )+3&=&\left ( -1 \right )+3 \\&2n&=&2 \\&\displaystyle \frac{2n}{2}&=&\displaystyle \frac{2}{2} \\ &\therefore n&=&1 \\ \end{array}$

3. $5a-1=4$

Solution:

Since the variable term $5a$ is subtracted by 1, then we add both sides by 1 then divide both sides both sides by 5.

$\begin{array}{c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c} &5a-1&=&4 \\ &5a-1+1&=&4+1 \\&5a&=&5 \\&\displaystyle \frac{5a}{5}&=&\displaystyle \frac{5}{5} \\ &\therefore a&=&1 \\ \end{array}$

4. $\displaystyle \frac{a}{3}+2=\left(-5\right)$

Solution:

Subtract 2 from both sides of the equation then multiply both sides by the inverse of $\frac {1}{3}$, which is 3.

$\begin{array}{c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c} &\displaystyle \frac{a}{3}+2&=&\left ( -5 \right ) \\ &\displaystyle \frac{a}{3}+2-2&=&\left ( -5 \right )-2 \\&\displaystyle \frac{a}{3}&=&-7 \\&\left ( 3 \right )\displaystyle \frac{a}{3}&=&-7\left ( 3 \right ) \\ &\therefore a&=&-21 \\ \end{array}$

Practice Exercises:

Solve for the unknown variable.

1. $2x-4x=-3$
2. $3w+\left(-7\right)=-19$
3. $-3y-3=12$
4. $\displaystyle \frac{z}{-3}+2=9$
5. $3q+3=-36$