FA14, HA17 - Solving linear equations

FA14, HA17 - Solving linear equations

An equation asks which value makes two expressions equal. Learn to undo operations, deal with brackets and fractions, and solve equations with the unknown on both sides. FA14 and HA17 also include using a supplied graph to estimate a solution, so you will connect equality with the crossing of two lines.

Inverse operations

What value makes 5x+25x+2 equal to 3x+153x+15? Trying whole numbers might miss the answer. Start with simpler equations to see how equality lets us find it exactly.

Solution of an equation

A solution is a value of the unknown that makes the original equation true. To solve an equation is to find its solution or solutions.

An equation is like a balanced pair of amounts. Adding or subtracting the same quantity on both sides keeps them equal. Multiplying or dividing both sides by the same non-zero number also keeps equality. These reversible steps produce an equivalent equation with the same solutions.

An inverse operation undoes an operation: subtraction undoes addition, and division undoes multiplication by a non-zero number. Work backwards from the last operation applied to the unknown. Dividing by zero is never allowed.

Worked example 1: undoing subtraction

Question. Solve x−7=14x-7=14.

Plan. Undo subtracting 77 by adding 77 to both sides.

x−7+7=14+7add the same amountx=21simplify both sides\begin{aligned} x-7+7&=14+7 &&\text{add the same amount}\\ x&=21 &&\text{simplify both sides} \end{aligned}

Answer. x=21x=21.

Check. In the original equation, 21−7=1421-7=14, as required.

Now the unknown is multiplied by 33 before 77 is subtracted. Undo those operations in reverse order.

Worked example 2: two inverse operations

Question. Solve 3x−7=143x-7=14.

Plan. Add 77 to both sides, then divide both sides by 33.

3x=21add seven to both sidesx=7divide both sides by three\begin{aligned} 3x&=21 &&\text{add seven to both sides}\\ x&=7 &&\text{divide both sides by three} \end{aligned}

Why this order? The final subtraction is undone before the earlier multiplication.

Answer. x=7x=7.

Check. 3×7−7=21−7=143\times7-7=21-7=14, so the original left side equals the right side.

Keep each equation as one complete equality on its own line.

Watch out

From 3x−7=143x-7=14, writing 3x=14−73x=14-7 changes only the right side in the wrong direction. Add 77 to both sides to obtain 3x=213x=21.

Try another two-step equation and check the value in the original.

Brackets and fractions

Brackets and fraction bars group expressions, so an operation on that group must include every term inside it.

You already know that 3(x+2)=3x+63(x+2)=3x+6. You can expand a bracket before solving, or first divide by its outside multiplier. With a numerical denominator, multiply both complete sides by that denominator to remove it. This is valid because the denominator is a fixed non-zero number.

Worked example 3: a bracket

Question. Solve 3(x+2)=213(x+2)=21.

Plan. Divide both sides by 33, then undo the addition inside the bracket.

x+2=7divide the whole equation by threex=5subtract two from both sides\begin{aligned} x+2&=7 &&\text{divide the whole equation by three}\\ x&=5 &&\text{subtract two from both sides} \end{aligned}

Answer. x=5x=5.

Check. 3(5+2)=3×7=213(5+2)=3\times7=21. Alternatively, expanding gives 3x+6=213x+6=21, which has the same solution.

Put a denominator under the bracketed expression. Clear this extra outer operation first.

Worked example 4: a bracket inside a fraction

Question. Solve 3(x+2)4=6\frac{3(x+2)}{4}=6.

Plan. Multiply both sides by 44, then solve the remaining bracket equation.

3(x+2)=24multiply both sides by fourx+2=8divide both sides by threex=6subtract two\begin{aligned} 3(x+2)&=24 &&\text{multiply both sides by four}\\ x+2&=8 &&\text{divide both sides by three}\\ x&=6 &&\text{subtract two} \end{aligned}

Answer. x=6x=6.

Check. 3(6+2)4=244=6\frac{3(6+2)}4=\frac{24}4=6, matching the right side of the original equation.

If two different denominators occur, choose a common multiple and multiply every term by it. For denominators 33 and 22, using 66 avoids introducing new fractions.

Worked example 5: two numerical denominators

Question. Solve x−23+x+12=7\frac{x-2}{3}+\frac{x+1}{2}=7.

Plan. Multiply both sides by 66, expand, then collect like terms.

2(x−2)+3(x+1)=42multiply every term by six2x−4+3x+3=42expand both brackets5x−1=42collect like terms5x=43add onex=435=8.6divide by five\begin{aligned} 2(x-2)+3(x+1)&=42 &&\text{multiply every term by six}\\ 2x-4+3x+3&=42 &&\text{expand both brackets}\\ 5x-1&=42 &&\text{collect like terms}\\ 5x&=43 &&\text{add one}\\ x&=\tfrac{43}{5}=8.6 &&\text{divide by five} \end{aligned}

Answer. x=8.6x=8.6.

Check. 8.6−23+8.6+12=2.2+4.8=7\frac{8.6-2}{3}+\frac{8.6+1}{2}=2.2+4.8=7. A solution need not be a whole number.

The multiplier applies to terms outside fractions too.

Watch out

Multiplying x3+2=5\frac{x}{3}+2=5 by 33 gives x+6=15x+6=15, not x+2=15x+2=15. Leaving the 22 unchanged breaks equality.

Clear both denominators in the next equation using one multiplier.

Continue this lesson.

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