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Solve linear equations with a reason for every step, including brackets and fractions, and rearrange a formula to make any letter its subject.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
In this lesson you learn to solve a linear equation as a chain of equivalent equations, naming the property that justifies each line, so that a solution is something you can defend rather than something you produced. You then apply the same moves to formulas, where the constants are letters, and rearrange $P = 2l + 2w$ or $F = \frac{9}{5}C + 32$ to make any letter the subject.
You already solve $3x + 4 = 19$ by undoing the addition and then the multiplication. You already expand $4(x - 3)$. What is new is being asked why each line follows from the one above — because a solution is a chain of equations that all have the same answer, and a step with no reason is a step that may have broken the chain.
Equivalent equations: equations with exactly the same solutions. Every legal step produces one.
Property of equality: a move applied to both sides — add, subtract, multiply or divide by the same thing (never divide by zero).
Distributive property: $a(b + c) = ab + ac$. It rewrites one side; it is not a move on the equation.
Solution: a value that makes both sides the same number. Substituting it back is how you check.
Solving is not a sequence of tricks; it is a chain of equations that all have the same solution set. Two kinds of move are allowed. The properties of equality change the equation but not its solutions: add, subtract, multiply or divide both sides by the same thing, provided you never multiply or divide by zero. The properties of arithmetic — distributive, commutative, associative — rewrite one side into an equal expression without touching the equation at all. A standard order works almost always: clear brackets, clear fractions, gather the variable on one side and the numbers on the other, then divide by the coefficient. Substituting the answer back into the original equation is the check, and it catches every arithmetic slip on the way.
Another way: diagram
A balance scale with $4(x - 3)$ in the left pan and $2x + 10$ in the right; below it the same scale with $2x$ removed from both pans, showing the beam still level.
Another way: story
Think of each line as a claim you are making about $x$, and the reason as the evidence. If somebody asked 'why does that line follow?' at any point, you should be able to name a property rather than say 'that is how it is done'.
"$4(x - 3)$ is $4x - 3$." The $4$ multiplies everything inside: $4x - 12$.
"I did it to one side, so it is fine." An equation is a balance. Subtracting $2x$ from the left only is not a step at all; it produces a different equation with a different answer.
"$\frac{x + 4}{3} = \frac{2x - 1}{5}$ means $x + 4 = 2x - 1$." Multiplying away the denominators means multiplying both sides by both of them: $5(x + 4) = 3(2x - 1)$.
Expand: $4x - 12 = 2x + 10$.
Distributive property — one side rewritten.
Subtract $2x$ from both sides: $2x - 12 = 10$.
Subtraction property of equality.
Add $12$ to both sides: $2x = 22$.
Addition property of equality.
Divide both sides by $2$: $x = 11$. Check: $4(11 - 3) = 32$ and $2(11) + 10 = 32$.
Division property, then the check.
Multiply both sides by $15$: $5(x + 4) = 3(2x - 1)$.
One multiplication clears both denominators.
Expand: $5x + 20 = 6x - 3$.
Distributive property, twice.
Subtract $5x$ and add $3$: $23 = x$, so $x = 23$.
Subtract $2x$ from both sides: $3x - 4 = 11$.
Add $4$, then divide by $3$: $3x = 15$, so $x = 5$.
Solve $2(x - 7) = 5x - 11$. Write one equation per line, ending with $x = $ your answer.
2(x - 7) = 5x - 11
Solve $9x + 2 = x - 54$ for $x$.
answer
Match each move to the property that justifies it.
| addition property of equality | division property of equality | distributive property | commutative property of addition | |
|---|---|---|---|---|
| Adding $8$ to both sides | ||||
| Dividing both sides by $3$ | ||||
| Rewriting $8(x + 5)$ as $8x + 40$ | ||||
| Replacing $4 + 5$ with $5 + 4$ |
Solve $\dfrac{x - 2}{5} = \dfrac{x - 6}{6}$ for $x$.
Answer:
One taxi firm charges $\$38$ to get in and $\$2$ a kilometre. A second charges $\$2$ to get in and $\$8$ a kilometre. After how many kilometres do the two journeys cost the same?
Answer:
You already solve $2w + 14 = 30$: subtract, then divide. A formula is the same job with letters where the numbers were. Nothing new has to be learned — what is hard is only that the answer is an expression rather than a number, so there is no arithmetic to reassure you at the end.
Literal equation: an equation whose constants are letters, such as $P = 2l + 2w$ or $I = Prt$.
Solve for a variable: rewrite the formula so that variable stands alone on one side and does not appear on the other.
Subject of the formula: the variable that has been left alone. In $P = 2l + 2w$, $P$ is the subject.
Reciprocal: to undo multiplying by $\frac{9}{5}$, multiply by $\frac{5}{9}$.
Solving for a letter uses exactly the moves you use on an equation with numbers, done in the same order: undo what is added or subtracted, then undo what multiplies or divides. The trick is to ask what is happening to the letter you want, working from the outside in. In $F = \frac{9}{5}C + 32$ the $C$ is multiplied by $\frac{9}{5}$ and then $32$ is added, so you subtract $32$ and then multiply by $\frac{5}{9}$: $C = \frac{5(F - 32)}{9}$. When the letter is only ever multiplied, as in $I = Prt$, one division does the whole job: $t = \frac{I}{Pr}$. And when the letter appears in a fraction, multiply both sides by the denominator first.
Another way: diagram
Two arrow chains for $F = \frac{9}{5}C + 32$: forwards, $C \to \times \frac{9}{5} \to +32 \to F$; and backwards underneath, $F \to -32 \to \times \frac{5}{9} \to C$.
Another way: story
Rearranging is unwrapping a parcel. The last thing done up is the first thing undone, so read the formula as a list of things that happened to the letter, then reverse the list.
"$P = 2l + 2w$, so $w = P - 2l$." The $2$ is still multiplying $w$. Undo the addition and the multiplication: $w = \frac{P - 2l}{2}$.
"$I = Prt$, so $t = \frac{I}{P} - r$." Nothing is added here, so nothing is subtracted. Divide by the whole product: $t = \frac{I}{Pr}$.
"There is no answer, it still has letters in it." An expression is the answer. Rearranging a formula is what turns one relationship into a recipe you can compute with.
What happens to $w$? It is multiplied by $2$, then $2l$ is added.
Read the formula as a chain.
Undo the addition: $P - 2l = 2w$.
Subtract $2l$ from both sides.
Undo the multiplication: $w = \frac{P - 2l}{2}$.
Divide both sides by $2$.
Nothing is added to $t$; it is multiplied by $P$ and by $r$.
So there is nothing to subtract.
Divide both sides by $Pr$: $t = \frac{I}{Pr}$.
One division undoes both multiplications.
Multiply both sides by $2$ to clear the fraction: $2A = bh$.
Divide both sides by $b$: $h = \frac{2A}{b}$.
A shape's perimeter is $P = 9l + 6w$. Solve for $w$.
Answer:
A volume is $V = 9abh$. Solve for $h$.
Answer:
A conversion rule is $F = \dfrac{8}{9}C + 24$. Solve for $C$.
Answer:
An area rule is $A = \dfrac{6bh}{2}$. Solve for $h$.
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Solve $5(x - 4) = 2x - 32$. Write one equation per line, ending with $x = $ your answer.
5(x - 4) = 2x - 32
A shape's perimeter is $P = 9l + 6w$. Solve for $w$.
Answer:
You can solve a linear equation with brackets or fractions and say why each step is allowed, and you can make any letter the subject of a formula. Explain why $P = 2l + 2w$ gives $w = \frac{P - 2l}{2}$ and not $w = P - 2l$.
8. Your turn: solve $5x - 4 = 2x + 11$, step 2
20. Your turn: the area of a triangle is $A = \dfrac{bh}{2}$. Solve for $h$, step 2