A17 - Solving linear equations

A17 - Solving linear equations

An equation asks which value makes two expressions equal. You will solve linear equations using inverse operations, brackets and unknowns on both sides, then read approximate solutions from a graph. Checking your answer in the starting equation turns a guess into a result you can trust.

Undoing operations to solve an equation

A number is multiplied by 33, then 77 is added. The result is 2222. What was the number? And would the same approach still work if the answer were negative or a fraction?

You already know

Inverse operations undo each other. You have used them to change the subject of a formula. A solution must make the left and right sides of an equation equal.

For the opening puzzle, subtract 77 from 2222 to get 1515, then divide by 33 to get 55. The order matters: undo the last operation first. Each operation must act on both sides, so equality is preserved.

Linear equation in one unknown

An equation that can be written as ax+b=0ax+b=0, where a≠0a\ne0 and xx is the unknown. Its highest power of the unknown is 11. For example, 3x+7=223x+7=22 is linear; x2+7=22x^2+7=22 is not.

An equation can have a fractional or negative solution. There is no rule saying that the unknown must be a positive whole number. The balancing and inverse-operation rules are not on the formulae sheet: learn them.

Worked example 1: two inverse operations

Question. Solve 3x+7=223x+7=22. Non-calculator.

Plan. Subtract 77 from both sides, then divide both sides by 33.

3x+7=223x=15subtract 7x=5divide by 3\begin{aligned} 3x+7&=22\\ 3x&=15&&\text{subtract }7\\ x&=5&&\text{divide by }3 \end{aligned}

Why this order? Addition was the last operation applied to 3x3x, so undo it before multiplication.

Answer. x=5x=5.

Check. 3(5)+7=15+7=223(5)+7=15+7=22.

Change only the right side from 2222 to 33. The same method now gives a negative fraction; that is a perfectly good solution.

Worked example 2: a fractional solution

Question. Solve 3x+7=33x+7=3. Non-calculator.

Plan. Use the same inverse operations and leave the answer as an exact fraction.

3x+7=33x=−4subtract 7x=−43divide by 3\begin{aligned} 3x+7&=3\\ 3x&=-4&&\text{subtract }7\\ x&=-\frac43&&\text{divide by }3 \end{aligned}

Why keep the fraction? A rounded decimal would only approximately satisfy the equation.

Answer. x=−43x=-\frac43.

Check. 3(−43)+7=−4+7=33(-\frac43)+7=-4+7=3.

A useful number trick hides the same algebra. Think of any number, double it, add 1010, halve the result, then subtract your starting number. You always get 55, because (2x+10)/2−x=x+5−x=5(2x+10)/2-x=x+5-x=5. A number trick becomes much less mysterious when its operations are visible.

Watch out

Wrong: 3x=123x=12, so x=12−3=9x=12-3=9. The 33 multiplies xx; subtraction does not undo multiplication. Right: x=12÷3=4x=12\div3=4.

Change the right side again and keep the exact value.

Solving equations with brackets

A bracket groups an expression into one amount. You can sometimes undo the outside multiplication first, but expanding is a reliable route when there are extra terms.

For 2(x+3)=142(x+3)=14, dividing by 22 first gives x+3=7x+3=7. For 2(x+3)+x=142(x+3)+x=14, expand first so that the two xx terms can be collected. Both approaches preserve equality; choose the shorter valid route.

Worked example 3: a positive bracket multiplier

Question. Solve 4(x−2)+3=154(x-2)+3=15. Non-calculator.

Plan. Expand, collect the constants, then isolate xx.

4x−8+3=15expand4x−5=15collect constants4x=20add 5x=5divide by 4\begin{aligned} 4x-8+3&=15&&\text{expand}\\ 4x-5&=15&&\text{collect constants}\\ 4x&=20&&\text{add }5\\ x&=5&&\text{divide by }4 \end{aligned}

Why expand every term? The multiplier 44 applies to the whole bracket, including −2-2.

Answer. x=5x=5.

Check. 4(5−2)+3=12+3=154(5-2)+3=12+3=15.

Change only the sign of the bracket multiplier. Multiplying a negative by a negative now makes a positive constant.

Worked example 4: a negative bracket multiplier

Question. Solve −4(x−2)+3=15-4(x-2)+3=15. Non-calculator.

Plan. Expand with signs intact, then solve the resulting linear equation.

−4x+8+3=15expand both terms−4x+11=15collect constants−4x=4subtract 11x=−1divide by −4\begin{aligned} -4x+8+3&=15&&\text{expand both terms}\\ -4x+11&=15&&\text{collect constants}\\ -4x&=4&&\text{subtract }11\\ x&=-1&&\text{divide by }-4 \end{aligned}

Why is the constant +8+8? It comes from (−4)(−2)(-4)(-2), not from (−4)(2)(-4)(2).

Answer. x=−1x=-1.

Check. −4(−1−2)+3=12+3=15-4(-1-2)+3=12+3=15.

You can check the expansion separately from the solution. That helps locate a sign error before it spreads through the remaining lines.

Watch out

Wrong: −3(x−5)=−3x−15-3(x-5)=-3x-15. Right: −3(x−5)=−3x+15-3(x-5)=-3x+15, because the two negative factors give a positive product.

Try another negative multiplier. The rule is the same, even though the solution changes.

Bringing unknown terms together

When the unknown is on both sides, remove one set of unknown terms before undoing the remaining operations.

For example, in 7x+2=3x+187x+2=3x+18, subtracting 3x3x from both sides leaves 4x+2=184x+2=18. You are removing the same amount from equal amounts. Keeping the unknown terms on the side with the larger coefficient often avoids a negative coefficient, but either side works.

Worked example 5: unknowns on both sides

Question. Solve 7x+2=3x+187x+2=3x+18. Non-calculator.

Plan. Subtract 3x3x, then remove the constant and divide.

7x+2=3x+184x+2=18subtract 3x4x=16subtract 2x=4divide by 4\begin{aligned} 7x+2&=3x+18\\ 4x+2&=18&&\text{subtract }3x\\ 4x&=16&&\text{subtract }2\\ x&=4&&\text{divide by }4 \end{aligned}

Why subtract 3x3x from both sides? It leaves just one variable term without changing which value solves the equation.

Answer. x=4x=4.

Check. The left side is 7(4)+2=307(4)+2=30; the right side is 3(4)+18=303(4)+18=30.

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