Explainer: Linear Equations with Variables on Both Sides

In this explainer, we will learn how to solve equations in which the variable is on each side of the equation.

It can happen that an equation involves the equality of two expressions, each of them containing a variable. For instance, consider the equation 3π‘₯+5=4π‘₯βˆ’2.

What does this equation tell us? We have an unknown number, called π‘₯, and the equation says that if we take three times this number and add five, we get the same number as when we take four times this number and take away two. We can represent this equation with a bar diagram.

We see that we need to remove the π‘₯-term from one side of the equation, so that the variable π‘₯ appears only on one side. The most natural way in this example is to take away 3π‘₯ from each side, as is shown in the next diagram.

The 3π‘₯ then disappears from the left-hand side (as it is removed), and we have on the right-hand side 4π‘₯βˆ’2βˆ’3π‘₯=π‘₯βˆ’2.

This leaves us with 5=π‘₯βˆ’2.

Now, we know that adding 2 to π‘₯βˆ’2 is π‘₯, so adding 2 to both sides will lead us to the solution: 5+2=π‘₯;

that is, π‘₯=7.

This step can be represented on our bar diagram as well.

We see that we have been actually using the additive and multiplicative properties of equalities, which state that if the same amount is added to (or subtracted from) both sides of the equality, or if both sides of the equation are multiplied (or divided) by the same number, then the equality is still true.

The strategy is therefore to get rid of the π‘₯-term on one side by subtracting it from both sides of the equation.

In the previous example, 3π‘₯+5=4π‘₯βˆ’2, we could have chosen to subtract 4π‘₯ from each side of the equation. This would have led to 3π‘₯+5βˆ’4π‘₯=βˆ’2,

which simplifies to βˆ’π‘₯+5=βˆ’2.

Subtracting 5 from each side gives βˆ’π‘₯=βˆ’2βˆ’5,

and so βˆ’π‘₯=βˆ’7.

Finding π‘₯ from βˆ’π‘₯ is just finding the opposite of βˆ’π‘₯, which is actually equivalent to multiplying by βˆ’1. The opposite of βˆ’7 is 7, so π‘₯=7.

We find, of course, the same result, only that it was slightly longer in the second case.

Example 1: Solving Equations with Variables on Each Side

Find the value of π‘₯: 4π‘₯+5=1+5π‘₯.

Answer

We want to solve the equation 4π‘₯+5=1+5π‘₯.

For this, we want all the π‘₯-terms to be on one side of the equation. To achieve this, we need to subtract from each side of the equation one of the π‘₯-terms. We can choose to subtract from each side the smaller π‘₯-term, 4π‘₯, so that it disappears from the left-hand side, and we find 5=1+5π‘₯βˆ’4π‘₯;

that is, 5=1+π‘₯.

Now, we simply need to subtract 1 from each side, so that we will have one π‘₯-term on one side and a number on the other side: 5βˆ’1=π‘₯,

which gives π‘₯=4.

We can check that our answer is correct by plugging this value into both expressions of our equation. We find 4β‹…4+5=16+5=21

and 1+5β‹…4=1+20=21.

We find the same value for both expressions, so we have, indeed, 4π‘₯+5=1+5π‘₯ when π‘₯=4.

Hence, the value of π‘₯ that makes the equation true is 4.

Example 2: Solving Equations with Variables on Each Side

Solve the equation βˆ’8π‘₯+13=βˆ’5π‘₯+26.

  1. π‘₯=βˆ’313
  2. π‘₯=βˆ’3
  3. π‘₯=βˆ’1
  4. π‘₯=βˆ’133

Answer

We want to solve the equation βˆ’8π‘₯+13=βˆ’5π‘₯+26.

For this, we want all the π‘₯-terms to be on one side of the equation. To achieve this, we subtract from each side of the equation one of the π‘₯-terms. We can choose to subtract from each side the smaller one, βˆ’8π‘₯, so that it disappears from the left-hand side, where it is now. As βˆ’8π‘₯ has a negative coefficient, subtracting it is actually adding 8π‘₯. Indeed, we have βˆ’(βˆ’8π‘₯)=+8π‘₯. We find 13=βˆ’5π‘₯+26+8π‘₯,

which is, once the π‘₯-terms have been combined on the right-hand side, 13=26+3π‘₯.

To get rid of the 26 on the right-hand side, we subtract it from each side so that we get 13βˆ’26=3π‘₯;

that is, βˆ’13=3π‘₯.

We have found the value of 3π‘₯, so dividing both sides by 3 will give us the value of π‘₯: π‘₯=βˆ’133.

We can check that our answer is correct by plugging this value into the original equation and checking that the equation holds true: βˆ’8β‹…ο€Όβˆ’133+13=βˆ’5β‹…ο€Όβˆ’133+26.

Example 3: Solving Equations with Variables on Each Side in a Geometry Context

Find the length of π‘‹π‘Œ.

Answer

We want to find the length of π‘‹π‘Œ. By observing the diagram, we see that 𝑋, π‘Œ, and 𝑍 are aligned. Therefore, we have π‘‹π‘Œ+π‘Œπ‘=𝑋𝑍.

We only know the length of π‘Œπ‘, while the other two lengths are given in a term of the unknown π‘Ž.

By substituting for the lengths of the segments as given on the diagram, we find 6π‘Ž+12=8π‘Žβˆ’2.

Solving for π‘Ž will allow us then to find the length of π‘‹π‘Œ, which is 6π‘Ž.

So, let us solve 6π‘Ž+12=8π‘Žβˆ’2. The π‘Ž-term in the left-hand side (6π‘Ž) can be removed from its side by subtracting it from both sides of the equation. This gives us 12=8π‘Žβˆ’2βˆ’6π‘Ž,

which simplifies to 12=2π‘Žβˆ’2.

We need now to remove the βˆ’2 from the side with the π‘Ž-term, so for this we add 2 to each side. We find 12+2=2π‘Ž;

that is, 2π‘Ž=14.

Dividing both sides by 2 now gives us the value of π‘Ž; namely, π‘Ž=14Γ·2=7.

We can quickly check that the value we have just found for π‘Ž is correct by plugging it into the original equation 6π‘Ž+12=8π‘Žβˆ’2. Is the equation true when π‘Ž=7?

The left-hand side is 6β‹…7+12=42+12=54,

and the right-hand side is 8β‹…7βˆ’2=56βˆ’2=54.

We find the same value for both expressions; hence, our result is correct; namely, π‘Ž=7.

Now, since the length of π‘‹π‘Œ is 6π‘Ž cm, we find that π‘‹π‘Œ=6β‹…7=42.cm

The length of π‘‹π‘Œ is 42 cm.

Example 4: Solving Equations with Variables on Each Side in a Geometry Context

Given that π‘šβˆ π΅=π‘šβˆ πΆ, use the information in the figure to find the perimeter of triangle 𝐴𝐡𝐢.

Answer

We want to find the perimeter of the triangle. Perimeter means β€œthe measure (distance) around”(β€œperi” means around), so it is the distance around the edge of a shape. Looking at the diagram, we see that the sides of the triangle are given in terms of an unknown, π‘₯.

Therefore, we need to use other information about the triangle to be able to find the value of π‘₯ and then find the perimeter of the triangle. We are given an important piece of information in the question; that is, π‘šβˆ π΅=π‘šβˆ πΆ. A triangle with two equal angles is isosceles. And we know that it implies that the two sides forming the third angle (here ∠𝐴) are equal. That means that 𝐴𝐡=𝐴𝐢, and so 5π‘₯βˆ’5=4π‘₯+1.

To have all π‘₯-terms on one side of the equation, we can subtract 4π‘₯ from both sides so that it is removed from the right-hand side. We get 5π‘₯βˆ’5βˆ’4π‘₯=1,

which simplifies to π‘₯βˆ’5=1.

Now, we need to add 5 to each side so that the π‘₯-term is on its own on one side of the equation. We find π‘₯=1+5=6.

Remember to do a quick check of the value found for π‘₯ at this point. When the value of 6 is plugged into 5π‘₯βˆ’5 and 4π‘₯+1, do we get the same value for both expressions? The answer is yes, we find a value of 25 for both.

Now that we have found the value of π‘₯ that makes an isosceles triangle, we find that 𝐴𝐡=5π‘₯βˆ’5=5β‹…6βˆ’5=25,𝐴𝐢=𝐴𝐡=25,cmcm and 𝐡𝐢=3π‘₯=3β‹…6=18.cm

The perimeter is then the sum of the three sides; that is, perimetercm=𝐴𝐡+𝐴𝐢+𝐡𝐢=25+25+18=68.

Our answer is that the perimeter of the triangle is 68 cm.

Example 5: Solving Equations with Variables on Each Side in a Real-Life Problem

The startup cost for a restaurant is $120,000, and each meal costs $10 for the restaurant to make. If each meal is then sold for $15, after how many meals does the restaurant break even?

Answer

Let us call π‘₯ the number of meals that need to be sold for the restaurant to break even. A business breaks even when the production costs are exactly balanced by the sales incomes.

We are going to write first an algebraic expression for the cost to produce π‘₯ meals. We need to translate mathematically β€œthe startup cost of a restaurant is $120,000, and each meal costs $10.” The total cost in dollars for π‘₯ meals is then 120,000+π‘₯β‹…10, which is written as 120,000+10π‘₯.

Then, we write an algebraic expression for the money earned in dollars when π‘₯ meals are sold. Since each meal is sold for $15, π‘₯ meals are sold for π‘₯β‹…15, which is simply written as 15π‘₯.

The restaurant breaks even when these two quantities are equal; that is, when 120,000+10π‘₯=15π‘₯.

We need to have both π‘₯-terms on the same side of the equation, so we subtract 10π‘₯ from both sides so that the π‘₯-term on the left-hand side is removed. We have 120,000=15π‘₯βˆ’10π‘₯;

that is, 120,000=5π‘₯.

We now divide both sides by 5 to find the value of π‘₯: 120,000Γ·5=π‘₯;

that is, π‘₯=24,000.

We have found that the restaurant needs to sell 24,000 meals in order to break even.

Key Points

  1. The additive and multiplicative properties of equalities state that if the same amount is added to (or subtracted from) both sides of the equality, or if both sides of the equation are multiplied (or divided) by the same number, then the equality is still true.
  2. These properties are used to solve an equation of the form π‘Žπ‘₯+𝑏=𝑐π‘₯+𝑑, so as to get an equation of the form 𝑝π‘₯=π‘ž.

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