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# 3.4: Equations with Variables on Both Sides

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Solve an equation with variables on both sides.
• Solve an equation with grouping symbols.
• Solve real-world problems using equations with variables on both sides.

## Solve an Equation with Variables on Both Sides

When the variable appears on both sides of the equation, we need to manipulate our equation such that all variables appear on one side, and only constants remain on the other.

Example 1

Dwayne was told by his chemistry teacher to measure the weight of an empty beaker using a balance. Dwayne found only one lb weights, and so devised the following way of balancing the scales.

Knowing that each weight is one lb, calculate the weight of one beaker.

Solution

We know that the system balances, so the weights on each side must be equal. We can write an algebraic expression based on this fact. The unknown quantity, the weight of the beaker, will be our $x$. We can see that on the left hand scale we have one beaker and four weights. On the right scale, we have four beakers and three weights. The balancing of the scales is analogous to the balancing of the following equation.

$x + 4 = 4x+ 3$

“One beaker plus 4 lbs equals 4 beakers plus 3 lbs”

To solve for the weight of the beaker, we want all the constants (numbers) on one side and all the variables (terms with $x$ in) on the other side. Look at the balance. There are more beakers on the right and more weights on the left. We will aim to end up with only $x$ terms (beakers) on the right, and only constants (weights) on the left.

$& \ x+4=4x+3 && \text{Subtract} \ 3 \ \text{from both sides}.\\& \underline{\ \quad -3= \;\;\;\;\; -3}\\& \ x+1=4x\\& \underline{ -x \quad \ =-x\;\;\;\;\;\;} && \ \text{Subtract} \ x \ \text{from both sides}.\\& \quad \ \ \ 1=3x\\& \quad \ \ \frac{1}{3} = \frac{\cancel{3}x}{\cancel{3}} & & \text{Divide both sides by} \ 3.\\& \quad \ \ x = \frac{1} {3}$

Answer The weight of the beaker is one-third of a pound.

We can do the same with the real objects as we have done with the equation. Our first action was to subtract three from both sides of the equals sign. On the balance, we could remove a certain number of weights or beakers from each scale. Because we remove the same number of weights from each side, we know the scales will still balance.

On the balance, we could remove three weights from each scale. This would leave one beaker and one weight on the left and four beakers on the right (in other words $x+1=4x$):

The next step we could do is remove one beaker from each scale leaving only one weight on the left and three beakers on the right and you will see our final equation: $1 = 3x$.

Looking at the balance, it is clear that the weight of the beaker is one-third of a pound.

Example 2

Sven was told to find the weight of an empty box with a balance. Sven found one lb weights and five lb weights. He placed two one lb weights in three of the boxes and with a fourth empty box found the following way of balancing the scales.

Knowing that small weights are one lb and big weights are five lbs, calculate the weight of one box.

We know that the system balances, so the weights on each side must be equal. We can write an algebraic expression based on this equality. The unknown quantity, the weight of each empty box, in pounds, will be our $x$. A box with two 1 lb weights in it weighs $(x+2)$. Here is the equation.

$& 3(x+2)=x + 3(5) & & \text{Distribute the} \ 3.\\& \ \ 3x+6=\cancel{x}+15\\& \ \underline{-x \quad \ \ = \cancel{-x}\;\;\;\;\;\;\;} & & \text{Subtract} \ x \ \text{from both sides}.\\& \quad 2x \cancel{+6}=15\\& \ \underline{\quad \ \ \cancel{-6}=-6\;\;\;\;\;\;\;\;} & & \text{Subtract} \ 6 \ \text{from both sides}.\\& \qquad 2x=9 & & \text{Divide both sides by} \ 2.\\& \qquad \ \ x = 4.5$

Solution

Each box weighs 4.5 lbs.

Multimedia Link To see more examples of solving equations with variables on both sides of the equation, see Khan Academy Solving Linear Equations 3

. This video has several more examples of solving equations and may help you practice the procedure of solving linear equations with variables on both sides of the equation.

## Solve an Equation with Grouping Symbols

When we have a number of like terms on one side of the equal sign we collect like terms then add them in order to solve for our variable. When we move variables from one side of the equation to the other we sometimes call it grouping symbols. Essentially we are doing exactly what we would do with the constants. We can add and subtract variable terms just as we would with numbers. In fractions, occasionally we will have to multiply and divide by variables in order to get them all on the numerator.

Example 3

Solve $3x + 4 = 5x$

Solution

This equation has $x$ on both sides. However, there is only a number term on the left. We will therefore move all the $x$ terms to the right of the equal sign leaving the constant on the left.

$& \ \ \cancel{3x}+4=5x & & \text{Subtract} \ 3x \ \text{from both sides}.\\& \underline{-\cancel{3x} \qquad \ -3x}\\& \qquad \quad 4=2x & & \text{Divide by} \ 2\\& \qquad \ \ \frac{4} {2} = \frac{2x}{2}$

Solution

$x=2$

Example 4

Solve $9x = 4-5x$

This time we will collect like terms ($x$ terms) on the left of the equal sign.

$& \ \ 9x=4\cancel{-5x}\\& \underline{+5x \quad \;\;\; \cancel{+5x}} & & \text{Add} \ 5x \ \text{to both sides}.\\& \ 14x = 4\\& \ 14x = 4 & & \text{Divide by} \ 14.\\& \frac{14x}{14} = \frac{4} {14}\\ & \ \quad x=\frac{2} {7}$

Solution

$x = \frac{2} {7}$

Example 5

Solve $3x + 2 = \frac{5x} {3}$

This equation has $x$ on both sides and a fraction. It is always easier to deal with equations that do not have fractions. The first thing we will do is get rid of the fraction.

$& \ \quad 3x + 2 = \frac{5x}{3}\\& 3(3x + 2) = 5x & & \text{Multiply both sides by} \ 3.\\& & & \text{Distribute the} \ 3.\\& \quad \cancel{9x}+6=5x\\& \underline{ - \cancel{9x} \qquad \ \ -9x} & & \text{Subtract} \ 9x \ \text{from both sides}:\\& \ \qquad \frac{6} {-4} = \frac{-4x}{-4}&& \text{Divide by} \ -4.\\& \ \qquad \frac{6}{-4}=x\\& \qquad -\frac{3}{2}=x$

Solution

$x = -1.5$

Example 6

Solve $7x + 2 = \frac{5x - 3} {6}$

Again we start by eliminating the fraction.

$& \quad \ 7x + 2 = \frac{5x - 3} {6}\\& 6(7x + 2) = \frac{5x - 3} {6} \cdot 6 & & \text{Multiply both sides by} \ 6.\\& 6(7x + 2) = 5x - 3 & & \text{Distribute the} \ 6.\\& \ 42x+12=\cancel{5x}-3 & & \text{Subtract} \ 5x \ \text{from both sides}:\\& \underline{-5x \quad \;\;\;\;\;\;-\cancel{5x}\;\;\;\;\;\;}\\& \ 37x+\cancel{12}=-3 & & \text{Subtract} \ 12 \ \text{from both sides}.\\ & \underline{ \qquad \ -\cancel{12}\quad -12\;\;\;\;\;\;}\\& \qquad \ 37x=-15 \\& \qquad \frac{37x}{37}=\frac{-15}{37} & & \text{Divide by} \ 37.$

Solution

$x = - \frac{15} {37}$

Example 7

Solve the following equation for $x$.

$\frac{14x} {(x + 3)} = 7$

The form of the left hand side of this equation is known as a rational function because it is the ratio of two other functions $(14x)$ and $(x+3)$. However, we wish simply to solve for $x$ so we start by eliminating the fraction. We do this as we have always done, by multiplying by the denominator.

$& \frac{14x} {(x + 3)} (x+3)= 7(x+3) & & \text{Multiply by} \ (x+3).\\& \qquad \qquad \quad 14x = 7(x + 3) & & \text{Distribute the} \ 7.\\& \qquad \qquad \quad 14x=\cancel{7x}+21\\& \underline{\qquad \quad \quad \ \ -7x=-\cancel{7x}\;\;\;\;\;\;\;} & & \text{Subtract} \ 7x \ \text{from both sides}.\\& \qquad \qquad \quad \ \ 7x=21\\& \qquad \qquad \quad \ \frac{7x}{7}=\frac{21}{7} & & \text{Divide both sides by} \ 7\\& \qquad \qquad \quad \ \ \ x=3$

## Solve Real-World Problems Using Equations with Variables on Both Sides

Build your skills in translating problems from words to equations. What is the equation asking? What is the unknown variable? What quantity will we use for our variable?

The text explains what is happening. Break it down into small, manageable chunks, and follow what is going on with our variable all the way through the problem.

## More on Ohm’s Law

The electrical current, $I$ (amps), passing through an electronic component varies directly with the applied voltage, $V$ (volts), according to the relationship:

$V=I \cdot R & & \text{where} \ R \ \text{is the resistance (measured in Ohms)}$

The resistance $R$ of a number of components wired in a series (one after the other) is given by: $R=r_1 +r_2 + r_3 + r_4 + \ldots$

Example 8

In an attempt to find the resistance of a new component, a scientist tests it in series with standard resistors. A fixed voltage causes a 4.8 amp current in a circuit made up from the new component plus a $15\Omega$ resistor in series. When the component is placed in a series circuit with a $50\Omega$ resistor the same voltage causes a 2.0 amp current to flow. Calculate the resistance of the new component.

This is a complex problem to translate, but once we convert the information into equations it is relatively straight forward to solve. Firstly we are trying to find the resistance of the new component (in Ohms, $\Omega$). This is our $x$. We do not know the voltage that is being used, but we can leave that as simple $V$. Our first situation has the unknown resistance plus $15\Omega$. The current is 4.8 amps. Substitute into the formula $V=I\cdot R$.

$V = 4.8(x + 15)$

Our second situation has the unknown resistance plus $50\Omega$. The current is 2.0 amps.

$V = 2(x+ 50)$

We know the voltage is fixed, so the $V$ in the first equation must equal the $V$ in the second. This means that:

$& 4.8(x + 15) = 2(x + 50) && \text{Distribute the constants}.\\& \ \ 4.8x+72=\cancel{2x}+100\\& \underline{\ -2x \qquad \ \ -\cancel{2x}\;\;\;\;\;\;\;\;\;\;} && \text{Subtract} \ 2x \ \text{from both sides}.\\& \quad 2.8x \cancel{+72}=100\\& \underline{\qquad \ \ \cancel{-72} \ \ -72 \;\;\;\;\;\;\;\;\;} & & \text{Subtract} \ 72 \ \text{from both sides}.\\& \ \qquad \ 2.8x=28 & & \text{Divide both sides by} \ 2.8.\\& \ \qquad \frac{2.8x}{2.8}=\frac{28}{2.8}\\& \qquad \quad \quad x=10$

Solution

The resistance of the component is $10 \ \Omega$.

## Lesson Summary

• If an unknown variable appears on both sides of an equation, distribute as necessary. Then subtract (or add) one term to both sides to simplify the equation to have the unknown on only one side.

## Review Questions

1. Solve the following equations for the unknown variable.
1. $3(x - 1) = 2(x + 3)$
2. $7(x + 20) = x + 5$
3. $9(x - 2) = 3x + 3$
4. $2 \left (a - \frac{1} {3}\right ) = \frac{2} {5} \left (a + \frac{2} {3}\right )$
5. $\frac{2} {7} \left (t + \frac{2} {3}\right ) = \frac{1} {5} \left (t - \frac{2} {3}\right )$
6. $\frac{1} {7} \left (v + \frac{1} {4}\right ) = 2 \left ( \frac{3v} {2}- \frac{5} {2}\right )$
7. $\frac{y-4} {11} = \frac{2}{5} \cdot \frac{2y + 1} {3}$
8. $\frac{z} {16} = \frac{2(3z + 1)} {9}$
9. $\frac{q} {16} + \frac {q} {6} = \frac{(3q + 1)} {9} + \frac{3} {2}$
2. Manoj and Tamar are arguing about how a number trick they heard goes. Tamar tells Andrew to think of a number, multiply it by five and subtract three from the result. Then Manoj tells Andrew to think of a number add five and multiply the result by three. Andrew says that whichever way he does the trick he gets the same answer. What was Andrew's number?
3. I have enough money to buy five regular priced CDs and have $6 left over. However all CDs are on sale today, for$4 less than usual. If I borrow $2, I can afford nine of them. How much are CDs on sale for today? 4. Five identical electronics components were connected in series. A fixed but unknown voltage placed across them caused a 2.3 amp current to flow. When two of the components were replaced with standard $10\Omega$ resistors, the current dropped to 1.9 amps. What is the resistance of each component? 5. Solve the following resistance problems. Assume the same voltage is applied to all circuits. 1. Three unknown resistors plus $20\Omega$ give the same current as one unknown resistor plus $70\Omega$. 2. One unknown resistor gives a current of 1.5 amps and a $15\Omega$ resistor gives a current of 3.0 amps. 3. Seven unknown resistors plus $18\Omega$ gives twice the current of two unknown resistors plus $150\Omega$. 4. Three unknown resistors plus $1.5\Omega$ gives a current of 3.6 amps and seven unknown resistors plus 7 $12\Omega$ resistors gives a current of 0.2 amps. ## Review Answers 1. $x = 9$ 2. $x = -22.5$ 3. $x = 3.5$ 4. $a = \frac{7}{12}$ 5. $t = -\frac{34}{9}$ 6. $v = \frac{141}{80}$ 7. $y = - \frac{82}{29}$ 8. $z = -\frac{32}{87}$ 9. $q = -\frac{232}{15}$ 1. 9 2.$7
3. $6.55\Omega$
1. unknown $= 25 \Omega$
2. unknown $= 30 \Omega$
3. unknown $= 94 \Omega$
4. unknown $= 1.213 \Omega$

Feb 22, 2012

Sep 28, 2014