<meta http-equiv="refresh" content="1; url=/nojavascript/"> Equations with Variables on Both Sides | CK-12 Foundation
You are reading an older version of this FlexBook® textbook: CK-12 Algebra I - Second Edition Go to the latest version.

3.4: Equations with Variables on Both Sides

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 a variable appears on both sides of the equation, we need to manipulate the equation so that all variable terms appear on one side, and only constants are left 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 them) on the other side. Since there are more beakers on the right and more weights on the left, we’ll try to move all the $x$ terms (beakers) to the right, and the constants (weights) to the left.

First we subtract 3 from both sides to get $x + 1 = 4x$.

Then we subtract $x$ from both sides to get $1 = 3x$.

Finally we divide by 3 to get $\frac{1}{3} = x$.

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

We can do the same with the real objects as we did with the equation. Just as we subtracted amounts from each side of the equation, we could remove a certain number of weights or beakers from each scale. Because we remove the same number of objects from each side, we know the scales will still balance.

First, 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 $x + 1 = 4x$ ):

Then we could remove one beaker from each scale, leaving only one weight on the left and three beakers on the right, to get $1 = 3x$:

Looking at the balance, it is clear that the weight of one 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 some 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 $(x + 2)$ pounds. Our equation, based on the picture, is $3(x + 2) = x +3(5)$.

Distributing the 3 and simplifying, we get $3x + 6 = x + 15$.

Subtracting $x$ from both sides, we get $2x + 6 = 15$.

Subtracting 6 from both sides, we get $2x = 9$.

And finally we can divide by 2 to get $x = \frac{9}{2}$, or $x = 4.5$.

Each box weighs 4.5 lbs.

To see more examples of solving equations with variables on both sides of the equation, see the Khan Academy video at http://www.youtube.com/watch?v=Zn-GbH2S0Dk.

Solve an Equation with Grouping Symbols

As you’ve seen, we can solve equations with variables on both sides even when some of the variables are in parentheses; we just have to get rid of the parentheses, and then we can start combining like terms. We use the same technique when dealing with fractions: first we multiply to get rid of the fractions, and then we can shuffle the terms around by adding and subtracting.

Example 3

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

Solution

The first thing we’ll do is get rid of the fraction. We can do this by multiplying both sides by 3, leaving $3(3x + 2) = 5x$.

Then we distribute to get rid of the parentheses, leaving $9x + 6 = 5x$.

We’ve already got all the constants on the left side, so we’ll move the variables to the right side by subtracting $9x$ from both sides. That leaves us with $6 = -4x$.

And finally, we divide by -4 to get $-\frac{3}{2} = x$, or $x = -1.5$.

Example 4

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

Solution

Again we start by eliminating the fraction. Multiplying both sides by 6 gives us $6(7x + 2) = 5x - 3$, and distributing gives us $42x + 12 = 5x - 3$.

Subtracting $5x$ from both sides gives us $37x + 12 = - 3$.

Subtracting 12 from both sides gives us $37x = -15$.

Finally, dividing by 37 gives us $x = -\frac{15}{37}$.

Example 5

Solve the following equation for $x$: $\frac{14x}{(x + 3)} = 7$

Solution

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 $(x + 3)$. But we can solve it just like any other equation involving fractions.

First we multiply both sides by $(x + 3)$ to get rid of the fraction. Now our equation is $14x = 7(x + 3)$.

Then we distribute: $14x = 7x + 21$.

Then subtract $7x$ from both sides: $7x = 21$.

And divide by 7: $x = 3$.

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

Here’s another chance to practice 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’s happening. Break it down into small, manageable chunks, and follow what’s going on with our variable all the way through the problem.

More on Ohm’s Law

Recall that the electrical current, $I$ (amps), passing through an electronic component varies directly with the applied voltage, $V$ (volts), according to the relationship $V = I \cdot R$ where $R$ is the resistance measured in Ohms $(\Omega)$.

The resistance $R$ of a number of components wired in a series (one after the other) is simply the sum of all the resistances of the individual components.

Example 6

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’s relatively straightforward to solve. First, we are trying to find the resistance of the new component (in Ohms, $\Omega$). This is our $x$. We don’t know the voltage that is being used, but we can leave that as a variable, $V$. Our first situation has a total resistance that equals the unknown resistance plus $15\Omega$. The current is 4.8 amps. Substituting into the formula $V = I \cdot R$, we get $V = 4.8(x + 15)$.

Our second situation has a total resistance that equals the unknown resistance plus $50\Omega$. The current is 2.0 amps. Substituting into the same equation, this time we get $V = 2(x + 50)$.

We know the voltage is fixed, so the $V$ in the first equation must equal the $V$ in the second. That means we can set the right-hand sides of the two equations equal to each other: $4.8(x + 15) = 2(x + 50)$. Then we can solve for $x$.

Distribute the constants first: $4.8x + 72 = 2x + 100$.

Subtract $2x$ from both sides: $2.8x + 72 = 100$.

Subtract 72 from both sides: $2.8x = 28$.

Divide by 2.8: $x = 10$.

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 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. $\frac{q}{16} + \frac{q}{6} = \frac{(3q + 1)}{9} + \frac{3}{2}$
10. $\frac{3}{x} = \frac{2}{x + 1}$
11. $\frac{5}{2 + p} = \frac{3}{p - 8}$
2. Manoj and Tamar are arguing about a number trick they heard. 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.
1. What was the number Andrew started with?
2. What was the result Andrew got both times?
3. Name another set of steps that would have resulted in the same answer if Andrew started with the same number.
3. Manoj and Tamar try to come up with a harder trick. Manoj tells Andrew to think of a number, double it, add six, and then divide the result by two. Tamar tells Andrew to think of a number, add five, triple the result, subtract six, and then divide the result by three.
1. Andrew tries the trick both ways and gets an answer of 10 each time. What number did he start out with?
2. He tries again and gets 2 both times. What number did he start out with?
3. Is there a number Andrew can start with that will not give him the same answer both ways?
4. Bonus: Name another set of steps that would give Andrew the same answer every time as he would get from Manoj’s and Tamar’s steps.
4. 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.
1. How much are CDs on sale for today?
2. How much would I have to borrow to afford nine of them if they weren’t on sale?
5. 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?
6. 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 seven $12\Omega$ resistors gives a current of 0.2 amps.

Date Created:

Feb 22, 2012

Sep 28, 2014
Files can only be attached to the latest version of None