C2.5; E2.5 - Forming and solving linear equations

C2.5; E2.5 - Forming and solving linear equations

A rectangle is 5 cm longer than it is wide, and its perimeter is 50 cm. How can one unknown describe both dimensions and lead to their values? C2.5 and E2.5 connect balanced equation solving with constructing, solving and checking models of lengths, costs and quantities.

Balanced operations

An equation says that its two sides have equal values. Imagine two balanced pans: removing the same amount from each pan keeps them balanced. Algebra uses the same principle to uncover an unknown value.

Solution of an equation

A solution is a value that makes the original equation true when substituted for the unknown. A linear equation in one unknown has that unknown only to the first power after simplification, with no unknown in a denominator.

Adding or subtracting the same expression on both sides preserves equality. Multiplying or dividing both sides by the same non-zero number also preserves it. Use inverse operations to undo what happened to the unknown, in reverse order. You cannot divide by zero.

For 3x+4=253x+4=25, the calculation multiplies xx by 3, then adds 4. Undo the addition first, then the multiplication. Write a complete new equation after each operation.

Worked example 1: undoing multiplication and addition

Question. Solve 3x+4=253x+4=25.

Plan. Subtract 4 from both sides, then divide both sides by 3.

3x+4=253x=21subtract 4 from both sidesx=7divide both sides by 3\begin{aligned} 3x+4&=25\\ 3x&=21 &&\text{subtract 4 from both sides}\\ x&=7 &&\text{divide both sides by 3} \end{aligned}

Answer. x=7x=7.

Check. Substitution in the original equation gives 3(7)+4=21+4=253(7)+4=21+4=25.

Change multiplication by 3 to division by 3. The last operation is still addition, so it is still undone first.

Worked example 2: undoing division and addition

Question. Solve x3+4=11\frac{x}{3}+4=11.

Plan. Subtract 4 from both sides, then multiply both sides by 3.

x3+4=11x3=7subtract 4x=21multiply both sides by 3\begin{aligned} \frac{x}{3}+4&=11\\ \frac{x}{3}&=7 &&\text{subtract 4}\\ x&=21 &&\text{multiply both sides by 3} \end{aligned}

Answer. x=21x=21.

Check. 213+4=7+4=11\frac{21}{3}+4=7+4=11, as required.

A solution can be negative or fractional. There is no rule that an unknown must be a positive whole number unless its context imposes that restriction.

Watch out

From 3x+4=253x+4=25, writing 3x=25+43x=25+4 changes the balance. Subtract 4 from both sides: 3x=25−43x=25-4. “Moving a term” is shorthand for an operation on both sides, not a separate rule.

Check the answer in the equation as given.

Unknowns and brackets

When the unknown appears on both sides, collect its terms on one side and the constants on the other. Choose a direction that keeps the arithmetic manageable. Expand any brackets before collecting, multiplying every term inside by the outside factor.

For example, subtracting 3x3x from both sides of 5x−3=3x+95x-3=3x+9 produces 2x−3=92x-3=9. The equality is preserved because exactly the same expression was subtracted from each side. Collecting unknown terms is another balanced operation.

Worked example 3: a negative solution

Question. Solve 5−2x=3x+105-2x=3x+10.

Plan. Collect the unknown terms on the right, then isolate xx.

5−2x=3x+105=5x+10add 2x to both sides−5=5xsubtract 10x=−1divide by 5\begin{aligned} 5-2x&=3x+10\\ 5&=5x+10 &&\text{add }2x\text{ to both sides}\\ -5&=5x &&\text{subtract 10}\\ x&=-1 &&\text{divide by 5} \end{aligned}

Answer. x=−1x=-1.

Check. The original left side is 5−2(−1)=75-2(-1)=7; the right side is 3(−1)+10=73(-1)+10=7.

The next right side contains a bracket. Expansion reveals the same sort of equation, but the solution is now fractional.

Worked example 4: brackets before collection

Question. Solve 5−2x=3(x+4)5-2x=3(x+4).

Plan. Expand the right side, then collect and divide as before.

5−2x=3x+12multiply both bracket terms by 35=5x+12add 2x−7=5xsubtract 12x=−75divide by 5\begin{aligned} 5-2x&=3x+12 &&\text{multiply both bracket terms by 3}\\ 5&=5x+12 &&\text{add }2x\\ -7&=5x &&\text{subtract 12}\\ x&=-\frac75 &&\text{divide by 5} \end{aligned}

Answer. x=−75x=-\frac75, or −1.4-1.4.

Check. In the original equation, the left side is 5−2(−1.4)=7.85-2(-1.4)=7.8, and the right side is 3(−1.4+4)=3(2.6)=7.83(-1.4+4)=3(2.6)=7.8.

Keep an exact fraction where it avoids premature rounding. Checking only your final simplified equation can miss an earlier expansion error: substitute into the original brackets.

Watch out

3(x+4)=3x+123(x+4)=3x+12, not 3x+43x+4. A negative outside factor also multiplies every term: −2(x−3)=−2x+6-2(x-3)=-2x+6.

Expand first, then apply balanced operations.

Forming equations from context

A model begins by deciding what the unknown measures. In the opening rectangle, write “let ww be the width in centimetres”. The length is then w+5w+5, not a second independent unknown, because its relationship to the width is known.

Equation model

An equation model represents a stated relationship between quantities using an equality. Define the unknown's meaning and unit, express related quantities in terms of it, then equate two expressions for the same quantity.

For a length problem, both sides must measure lengths in the same units. For a cost problem, the item costs and the total must use the same currency unit. Count a fixed charge once and a per-item charge once for each item.

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