Distribution, Rational Coefficients and Error Analysis | Grade 8 Math | LevelKite

You can already solve an equation such as $5x + 2 = 3x + 10$ by collecting the parts with $x$ on one side and the plain numbers on the other. The letter $x$ is a variable: it stands for one unknown number, which may be any real number (any number on the number line). A solution is a value of $x$ that makes the equation true, which means both sides have the same value.

In this lesson the equations arrive in disguise, with parentheses, fractions or decimals. Your job is to remove the disguise without changing the solutions, and then to spot where someone else's solving went wrong.

Start with parentheses. The expression $3(x + 4)$ means 3 groups of the entire quantity $x + 4$ . Three groups hold three copies of $x$ and three copies of $4$ , so $3(x + 4)$ has the same value as $3x + 12$ whatever number $x$ is. Two expressions that have the same value for every permitted value of the variable (here, every real number) are called equivalent expressions. Swapping one for the other never changes which values make an equation true.

This is the distributive property: when a number multiplies a quantity in parentheses, it multiplies every term inside. A term is one of the parts being added in an expression. A subtracted part counts as a term with a negative sign, so the terms of $x - 11$ are $x$ and $-11$ . In symbols, for any numbers $a$ , $b$ and $c$ :

a(b + c) = ab + ac

Key concept

Two kinds of moves keep the solutions the same

Solving an equation means replacing it with simpler equations that have exactly the same solutions. Two kinds of moves are guaranteed to do that:

Rewrite one side as an equivalent expression, for example by distributing to every term inside parentheses.
Do the same operation to each entire side: add or subtract the same term, or multiply or divide by the same number, as long as that number is not zero.

A line that changes only part of a side, or treats the two sides differently, may have different solutions. So always check your answer in the original equation, not in a later line.

Worked example

Parentheses on both sides

Solve $3(x + 4) = -2(x - 11)$ .

Rewrite the left side. The 3 multiplies the entire quantity x + 4, so multiply each term inside: 3 times x is 3x, and 3 times 4 is 12. The new left side is equivalent to the old one, so the solutions do not change.
$3x + 12 = -2(x - 11)$
Rewrite the right side the same way. The terms inside are x and negative 11. Multiply each by negative 2: negative 2 times x is negative 2x, and negative 2 times negative 11 is positive 22, because a negative number times a negative number is positive. This rewrite is also equivalent.
$3x + 12 = -2x + 22$
Add 2x to both sides so the variable terms are together on the left. On the left, 3x plus 2x is 5x. On the right, negative 2x plus 2x is zero. Adding the same term to both sides keeps the same solutions.
$5x + 12 = 22$
Subtract 12 from both sides to leave the variable term alone on the left. On the right, 22 minus 12 is 10. The same number is subtracted from both sides, so the solutions stay the same.
$5x = 10$
Divide both sides by 5, the number multiplying x. This is allowed because 5 is not zero. On the right, 10 divided by 5 is 2.
$x = 2$
Check in the original equation by replacing x with 2. The left side is 3 times 6, which is 18. The right side is negative 2 times negative 9, which is also 18. Both sides match, so the equation is true.
$3(2 + 4) = 18 \quad \text{and} \quad -2(2 - 11) = 18$

Answer

The solution is $x = 2$ . It makes both sides of the original equation equal $18$ .

Now for fractions and decimals. A rational number is a number that can be written as a fraction of two integers (whole numbers and their opposites) with a nonzero denominator. So $\frac{2}{3}$ , $0.5$ and $-4$ are all rational. An equation such as $\frac{1}{2}x + 3 = \frac{2}{3}x + 1$ has rational coefficients on both sides.

You could solve with the fractions as they are, but there is a tidier route. If two sides are equal, then 6 times one side equals 6 times the other, and dividing by 6 takes you back. So multiplying each entire side by the same nonzero number neither adds nor removes solutions. Choose a number that turns every fraction or decimal into an integer. For fractions, use a common denominator: a positive number that every denominator divides evenly. The smallest one keeps the arithmetic light. For decimals, use $10$ when they go to tenths and $100$ when they go to hundredths.

Choosing a number that clears fractions or decimals

Fractions or decimals in the equation	Multiply both sides by	Reason
$\frac{1}{2}$ and $\frac{2}{3}$	$6$	$6$ is the smallest positive number that $2$ and $3$ both divide evenly
$\frac{3}{4}$ and $\frac{5}{6}$	$12$	$12$ is the smallest positive number that $4$ and $6$ both divide evenly
$0.5$ and $0.2$	$10$	both go to tenths: $10 \times 0.5 = 5$ and $10 \times 0.2 = 2$
$0.25$ and $0.4$	$100$	the longer one goes to hundredths: $100 \times 0.25 = 25$ and $100 \times 0.4 = 40$

Worked example

Fractions on both sides

Solve $\frac{1}{2}x + 3 = \frac{2}{3}x + 1$ .

The denominators are 2 and 3, and the smallest positive number both divide evenly is 6. Multiply each entire side by 6. Because 6 is not zero, the solutions stay the same. The parentheses show that the entire side is multiplied.
$6\left(\frac{1}{2}x + 3\right) = 6\left(\frac{2}{3}x + 1\right)$
Distribute the 6 to every term on each side, which gives equivalent expressions. On the left, 6 times one half of x is 3x, and 6 times 3 is 18. On the right, 6 times two thirds of x is 4x, and 6 times 1 is 6.
$3x + 18 = 4x + 6$
Subtract 3x from both sides, the same term on each side, so the variable terms are together on the right. On the left, 3x minus 3x is zero. On the right, 4x minus 3x is x.
$18 = x + 6$
Subtract 6 from both sides. On the left, 18 minus 6 is 12. The variable is now alone, and 12 = x says the same thing as x = 12.
$12 = x$
Check in the original equation by replacing x with 12. The left side is half of 12 plus 3, which is 6 + 3 = 9. The right side is two thirds of 12 plus 1, which is 8 + 1 = 9. Both sides match.
$\frac{1}{2}(12) + 3 = 9 \quad \text{and} \quad \frac{2}{3}(12) + 1 = 9$

Answer

The solution is $x = 12$ . It makes both sides of the original equation equal $9$ .

Worked example

Parentheses and decimals together

Solve $0.5(x + 4) = 0.2x + 3$ .

Distribute first to remove the parentheses. Multiply each term inside by 0.5: 0.5 times x is 0.5x, and 0.5 times 4 is 2. The left side is now an equivalent expression.
$0.5x + 2 = 0.2x + 3$
The decimals go to tenths, so multiply each entire side by 10, which is not zero. Every term is multiplied: 10 times 0.5x is 5x, 10 times 2 is 20, 10 times 0.2x is 2x, and 10 times 3 is 30.
$5x + 20 = 2x + 30$
Subtract 2x from both sides to collect the variable terms on the left. On the left, 5x minus 2x is 3x.
$3x + 20 = 30$
Subtract 20 from both sides. On the right, 30 minus 20 is 10.
$3x = 10$
Divide both sides by 3, which is not zero. Because 10 divided by 3 is not a whole number, write the result as an exact fraction.
$x = \frac{10}{3}$
Check in the original equation. Left side: 10/3 plus 4 is 22/3, and half of 22/3 is 11/3. Right side: 0.2 times 10/3 is 2/3, and 2/3 plus 3 is 11/3. Both sides match.
$0.5\left(\frac{10}{3} + 4\right) = \frac{11}{3} \quad \text{and} \quad 0.2\left(\frac{10}{3}\right) + 3 = \frac{11}{3}$

Answer

The solution is $x = \frac{10}{3}$ , an exact value. It makes both sides of the original equation equal $\frac{11}{3}$ .

The same two kinds of moves help you find mistakes in anyone's work, including your own. Treat the work like an audit, a careful line-by-line inspection:

Test the final answer in the original equation. If the two sides do not match, the answer is not a solution, so some line of the work contains an error.
Read the work one line at a time. For each line, name the move. Was it a rewrite that gives an equivalent expression, or the same operation on each entire side? Was the arithmetic right?
Stop at the first line that fails. Correct it, then redo every line after it, because the later lines were built on the mistake.
Check the new answer in the original equation.

The table shows what each move needs and what a slip can look like.

What each move needs in order to be valid

Move	Valid only when	A slip looks like
Distribute	the number outside multiplies every term inside, with the correct sign	$-3(x - 2)$ rewritten as $-3x - 6$ instead of $-3x + 6$
Clear fractions or decimals	every term on both sides is multiplied by the same nonzero number	$0.5x + 2$ times $10$ written as $5x + 2$ instead of $5x + 20$
Add or subtract a term	the same term is added to both sides, or subtracted from both sides	$7x + 1 = 2x + 16$ turned into $9x + 1 = 16$ instead of $5x + 1 = 16$
Divide by a coefficient	each entire side is divided by the same nonzero number	$2x - 8 = 12$ turned into $x - 8 = 6$ instead of $x - 4 = 6$

Worked example

Find and fix the error

Dana solved $2(3x - 4) = 4x + 12$ and got $x = 2$ . Her lines were (1) $6x - 8 = 4x + 12$ , (2) $10x - 8 = 12$ , (3) $10x = 20$ and (4) $x = 2$ . Find her error and correct it.

Test Dana's answer in the original equation by replacing x with 2. The left side is 2 times (6 minus 4), which is 4. The right side is 8 plus 12, which is 20. The sides do not match, so x = 2 is not a solution and some line contains an error.
$2(3 \cdot 2 - 4) = 4 \quad \text{but} \quad 4 \cdot 2 + 12 = 20$
Audit line 1. The move is distributing the 2 on the left side: 2 times 3x is 6x, and 2 times negative 4 is negative 8. Every term inside was multiplied and the right side was left alone, so line 1 is valid.
$2(3x - 4) = 6x - 8$
Audit line 2. To clear 4x from the right side, the same term must be subtracted from both sides. Dana's right side lost 4x, but her left side grew from 6x to 10x. She subtracted on the right and added on the left, so this line is the error. Correct it: 6x minus 4x is 2x.
$2x - 8 = 12$
Redo the later lines, because Dana's lines 3 and 4 were built on the wrong line. Add 8 to both sides: on the left only 2x remains, and on the right 12 plus 8 is 20.
$2x = 20$
Divide both sides by 2, which is not zero. On the right, 20 divided by 2 is 10.
$x = 10$
Check the new answer in the original equation by replacing x with 10. The left side is 2 times (30 minus 4), which is 52. The right side is 40 plus 12, which is 52. Both sides match.
$2(3 \cdot 10 - 4) = 52 \quad \text{and} \quad 4 \cdot 10 + 12 = 52$

Answer

Dana's error is in line 2: she added $4x$ on the left but subtracted it on the right. Subtracting $4x$ from both sides gives the correct solution, $x = 10$ .

Here is the idea to keep. Parentheses, fractions and decimals do not need new rules. The distributive property removes parentheses, and multiplying every term on both sides by one well-chosen nonzero number removes fractions and decimals. After that, you are back to an equation you already know how to solve. Every line must come from a valid move, and the check in the original equation is what catches a line that did not. Dana's $x = 2$ made her own later lines true; only the original equation exposed the error.

Next, you will use these skills to solve word problems that lead to equations with variables on both sides.