# Posts Tagged: "equations"

## 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. $3q+3=-36$

## One-Step Equation Word Problems

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One–step equation word problem is a simple word problem involving only addition or subtraction of a constant to an unknown value and either multiplying or dividing a certain number to an unknown quantity.

## One-Step Equations Containing Fractions

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After defining one–step equations and learning how to solve those if integers or decimals are involved, now let us learn how to solve one–step equations containing fractions. The rules for solving one–step equations containing integers or decimals still hold true for one–step equations involving fractions.

Still the underlying purpose is to separate the variable and the constant part, by placing the variable on the left side and the constant on the right side of the equation. To do so, perform the opposite of whatever operation is involved in the original equation or by doing the inverse of what is being done to the variable. Since fractions are involved, please note the rules for performing operations with fractions.

## One-Step Equations Containing Decimals

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After defining one–step equations and learning how to solve those if they contained integers, now let us learn how to solve one–step equations containing decimals. The rules for solving one–step equations containing integers still hold true for one–step equations involving decimals. Likewise, the underlying objective is to separate the variable and the constant part, by placing the variable on the left side and the constant on the right side of the equation. This can be done by doing the opposite of whatever operation is involved in the original equation or by performing the inverse of what is being done to the variable. Since decimals are involved, please note the rules for performing operations with decimals.

## One–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 or more steps before arriving at its solutions. Equations that require only one step or operation are called one-step equations. To solve one–step equation, we simply do the opposite of whatever operation is involved or to do the inverse of what has done to the variable.