# Solving for Variables

Solving equations with one or more variables is a crucial skill in math and science. The concept of solving equations can be applied to many real-world situations, such as calculating the amount of time it will take to travel $$300$$ miles at a speed of $$50$$ miles per hour. The equation $$D=rt$$ can be used to solve this problem by isolating the variable $$t$$. An equation is simply a mathematical representation for two things that are equal. This equality allows us to solve for unknown pieces of the equation. These unknown values are called variables. There are three steps required to solve most equations:

Solving for Variables Sample Questions

Step 1: Simplify both sides of the equation.
Step 2: Move all of the parts containing the variable you are solving for to the same side of the equation.
Step 3:. Isolate the variable using inverse operations.

The main objective when solving an equation is to isolate the variable. When the variable is on its own, it reveals the solution. The solution of an equation is a value that makes the equation balanced. For example, in the equation $$10x+30=90$$, the solution for $$x$$ is $$6$$ because when $$6$$ is multiplied by $$10$$, and then added to $$30$$, the result is $$90$$, creating a balanced equation.

Many equations will have only one variable, as in the previous example. However, some equations will have more than one variable. Let’s look at a few examples of equations with more than one variable.

The equation $$3x+2y=8$$ contains two variables. If the value of $$x$$ is $$2$$, what will the value of $$y$$ be? The first step in this example is to plug in $$2$$ for $$x$$. $$3x+2y=8$$ becomes $$3(2)+2y=8$$. Now multiply $$3\times2$$ so the equation becomes $$6+2y=8$$. At this point, there is only one variable in the equation. Now the goal is to isolate the variable $$y$$. This can be done by “undoing” the operations that are affecting the $$y$$. In order to “undo” a positive $$6$$ on the left side of the equation, we need to subtract $$6$$. This needs to be done to both sides of the equation in order to keep it balanced. $$6+2y=8$$ becomes $$2y=2$$. Now we are only one step away from knowing the value of $$y$$. $$y$$ is currently multiplied by $$2$$, so to “undo” this operation we need to divide both sides by $$2$$. Now the equation shows $$y=\frac{2}{2}$$ or $$y=1$$. To check that we have solved correctly, take this value for $$y=1$$, and plug it back into the equation to see if it is truly balanced: $$6+2(1)=8$$.

“Undoing” operations is referred to as inverse operations. Inverse operations are like opposite operations. The inverse of adding $$5$$ is subtracting $$5$$, and the inverse of multiplying by $$8$$ is dividing by $$8$$. Inverse operations are crucial for solving one- and two-step equations.

Let’s look at another example of an equation with multiple variables. This equation is often seen in the world of geometry. The equation is used to find the area of a circle, and it states that $$A=πr^2$$. This equation has two variables, $$A$$, and $$r$$. Let’s use the formula to determine the radius of a circle with an area of $$150$$ square inches.

When the values are plugged into the equation, $$A=πr^2$$ becomes $$150=πr^2$$. From here we can isolate the variable $$r$$ by dividing both sides by $$π$$, and then finding the square root of this. $$150$$ divided by $$π$$ is $$47.77$$, and the square root of this is approximately $$6.9$$. This means that the radius of the circle is approximately $$6.9$$ inches.

Formulas involving speed, area, or radius are only a few examples of the value of solving for variables. This skill is widely used in the world of math and science, so it is important to become familiar with the process. Remember, the main objective is to isolate the variable, which usually involves using inverse operations.

## Solving for Variables Sample Questions

Here are a few sample questions going over solving for variables.

Question #1:

Solve the equation for $$x$$.

$$4x-15=1$$

$$x=7$$
$$x=2$$
$$x=4$$
$$x=9$$

The goal of solving an equation is to isolate the variable. In this example, the $$x$$ is multiplied by $$4$$, and the $$15$$ is subtracted. In order to get the $$x$$ on its own we need to “undo” these operations. The inverse of subtracting $$15$$ is adding $$15$$. When $$15$$ is added to both sides of the equation, $$4x-15=1$$ becomes $$4x=16$$. To “undo” multiplication of $$4$$, we need to divide both sides of the equation by $$4$$. $$4x=16$$ becomes $$x=4$$. This means that if $$4$$ is plugged into the original equation for $$x$$, the equation will be balanced: $$4(4)-15=1$$.

Question #2:

Solve for the variable $$y$$, if $$x=3$$.

$$4x+2y=8$$

$$y=-5$$
$$y=-2$$
$$y=-9$$
$$y=-1$$

When $$3$$ is plugged into the equation for $$x$$, $$4x+2y=8$$ becomes $$4(3)+2y=8$$, which simplifies to $$12+2y=8$$. Now the objective is to isolate the variable $$y$$. In order to get y on its own, we need to “undo” the operations that are affecting $$y$$. First subtract $$12$$ from both sides of the equation, and then divide both sides by $$2$$. $$12+2y=8$$ becomes $$y=-2$$. To check that this is correct, plug in $$-2$$ for $$y$$ in the original equation. If the equation is balanced, then the answer is correct: $$12+2(-2)=8$$.

Question #3:

A phone company charges new customers a one-time installation fee of $$45$$, and then $$29$$ per month for service, which corresponds with the equation $$y=29x+45$$. If the customer doesn’t want to pay this company more than $$400$$, how many months will he be a customer?

$$9$$ months

$$12$$ months

$$18$$ months

$$24$$ months

If the customer doesn’t want to pay this company more than $$400$$, then $$y$$ cannot be greater than $$400$$. To solve this problem, set $$y$$ equal to $$400$$ and solve for $$x$$. This gives us the equation $$400=29x+45$$. To isolate the variable $$x$$, first subtract $$45$$ from both sides to “undo” the addition. This gives the equation $$355=29x$$. Next, divide both sides by $$29$$ to “undo” the multiplication. This results in $$x\approx12.24$$. So in order to not pay the company more than $$400$$, the customer has to cancel his membership after $$12$$ months.

Question #4:

Dan wants to build a small wooden box that has a volume of $$216$$ cubic centimeters. If the length and the width are $$6\text{ cm}$$ and $$9\text{ cm}$$, how tall will the box be?
Volume Formula for rectangular prisms: $$V=l\times w\times h$$

$$h=8\text{ cm}$$
$$h=12\text{ cm}$$
$$h=6\text{ cm}$$
$$h=4\text{ cm}$$

When the values are plugged into the formula, $$V=l\times w\times h$$ becomes $$216=6\times9\times h$$, where $$h$$ is the unknown variable that represents height. In order to “undo” the multiplication of $$9$$ and $$6$$, both sides of the equation need to be divided by $$9$$ and $$6$$. When the right side of the equation is divided by $$9$$ and $$6$$, $$h$$ is left on its own. When the left side of the equation is divided by $$9$$ and $$6$$ the result is $$4$$. Therefore, the height of the box is $$4\text{ cm}$$.

Question #5:

The rate formula is $$D=rt$$ where $$D$$ represents distance, $$r$$ represents rate, and $$t$$ represents time. Use this formula to determine how long it would take to travel $$200$$ miles at a rate of $$40$$ miles per hour.

$$40\text{ minutes}$$
$$5\text{ hours}$$
$$4\text{ hours}$$
$$30\text{ minutes}$$
When the values are plugged into the rate formula, $$D=rt$$ becomes $$200=40t$$. In order to solve for the variable $$t$$, we need to divide both sides of the equation by $$40$$ in order to “undo” the multiplication by $$40$$. $$200=40t$$ becomes $$5=t$$ which represents $$5$$ hours.