6.03a - Solving linear equations

6.03a - Solving linear equations

An equation says that two expressions have the same value. You will keep that balance while undoing operations, collecting unknown terms, and handling brackets and fractions. You will also turn a word problem into an equation and decide what its solution means in the original situation.

Undoing operations

Five identical notebooks and a £7 delivery charge cost £32 altogether. How much does one notebook cost?

If each notebook costs xx pounds, the equation is 5x+7=325x+7=32. Removing the delivery charge leaves £25 for five notebooks, so each costs £5. Algebra records those same inverse steps.

Equation and solution

An equation states that two expressions are equal. A solution is a value of the unknown that makes that statement true. For example, x=5x=5 solves 5x+7=325x+7=32, because 5×5+7=325\times5+7=32.

Adding or subtracting the same amount on both sides preserves equality. Multiplying or dividing both sides by the same nonzero number also preserves equality. To undo a sequence of operations, work backwards: undo the addition or subtraction before undoing the multiplication. Dividing by zero is never a valid step.

Worked example 1: reversing two operations

Question. Solve 5x+7=325x+7=32.

Plan. Subtract 77 from both sides, then divide both sides by 55.

5x+7=32given equation5x=25subtract 7x=5divide by 5\begin{aligned} 5x+7&=32 && \text{given equation} \\ 5x&=25 && \text{subtract 7} \\ x&=5 && \text{divide by 5} \end{aligned}

Answer. x=5x=5.

Check. Substitute into the original: 5(5)+7=25+7=325(5)+7=25+7=32, matching the right side.

Change only the sign of the constant on the left. The inverse operation changes too, and the solution need not be a whole number.

Worked example 2: undoing subtraction

Question. Solve 5x−7=325x-7=32.

Plan. Add 77 to both sides before dividing by 55.

5x−7=32given equation5x=39add 7x=395=7.8divide by 5\begin{aligned} 5x-7&=32 && \text{given equation} \\ 5x&=39 && \text{add 7} \\ x&=\frac{39}{5}=7.8 && \text{divide by 5} \end{aligned}

Answer. x=7.8x=7.8.

Check. 5(7.8)−7=39−7=325(7.8)-7=39-7=32. A fractional or decimal solution is allowed unless the context restricts the unknown.

The word “move” can hide what is happening. You are not simply carrying a term across an equals sign: you are doing the same operation to both sides.

Watch out

From 5x+7=325x+7=32, writing 5x=32+75x=32+7 uses the wrong inverse. Subtract 77 from both sides to remove the +7+7.

Solve, then substitute to check your result.

Unknowns on both sides

When the unknown appears on both sides, first collect its terms on one side by the same balancing operations.

For example, two hire plans cost 7h+47h+4 pounds and 3h+243h+24 pounds for hh hours. Equal cost means 7h+4=3h+247h+4=3h+24. Subtracting the shared 3h3h leaves the extra 4h4h to balance the difference in fixed charges. This is why collecting terms works.

Worked example 3: collecting the unknown terms

Question. Solve 7x+4=3x+247x+4=3x+24.

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

7x+4=3x+24given equation4x+4=24subtract 3x4x=20subtract 4x=5divide by 4\begin{aligned} 7x+4&=3x+24 && \text{given equation} \\ 4x+4&=24 && \text{subtract }3x \\ 4x&=20 && \text{subtract 4} \\ x&=5 && \text{divide by 4} \end{aligned}

Answer. x=5x=5. In the hire-plan example, the costs are equal after five hours.

Check. The original left side is 35+4=3935+4=39; the right side is 15+24=3915+24=39.

Change only the sign of the constant on the right. A negative answer is just as valid for an equation without a restrictive context.

Worked example 4: a negative solution

Question. Solve 7x+4=3x−247x+4=3x-24.

Plan. Collect the unknown terms in the same way.

7x+4=3x−24given equation4x+4=−24subtract 3x4x=−28subtract 4x=−7divide by 4\begin{aligned} 7x+4&=3x-24 && \text{given equation} \\ 4x+4&=-24 && \text{subtract }3x \\ 4x&=-28 && \text{subtract 4} \\ x&=-7 && \text{divide by 4} \end{aligned}

Answer. x=−7x=-7.

Check. Left side: −49+4=−45-49+4=-45. Right side: −21−24=−45-21-24=-45. Both sides agree.

You could instead subtract 7x7x and obtain a negative coefficient. That is equally valid if every operation is applied to both sides.

Sometimes all unknown terms cancel. The equation 3x+2=3x+23x+2=3x+2 reduces to 2=22=2 and is true for every real xx. But 3x+2=3x+53x+2=3x+5 reduces to 2=52=5, so it has no solution. Do not divide by a zero coefficient to force a single answer.

Key idea

An ordinary linear equation with a nonzero final coefficient has one solution. Complete cancellation can instead reveal an identity or an impossible statement.

Keep the sign of the final constant in this equation.

Equations with brackets and fractions

Brackets and numerical fraction denominators change the preparation, but equality is still preserved by balancing operations.

Suppose three packs each contain x−2x-2 counters and their total equals x+8x+8 counters. This gives 3(x−2)=x+83(x-2)=x+8. Expand to get 3x−6=x+83x-6=x+8, then 2x=142x=14, so x=7x=7. Each pack has five counters, and 3×5=15=7+83\times5=15=7+8.

For fractions, multiply every term on both sides by a common multiple of the numerical denominators. This clears the fraction bars. Keep brackets around each whole numerator while multiplying; the operation applies to all of it.

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