Solving One- and Two-Step Equations

Undo the addition using the same subtraction on both sides.

Use the operation that undoes subtracting 8.

Four equal amounts total 28; find one of those amounts.

What operation undoes dividing by 3?

The unknown can be negative.

Multiply both sides by the denominator and keep the negative sign.

1. Multiply both sides by 4: a=-3\times4=-12.
2. Check: -12\div4=-3.

Undo the addition first, then undo the multiplication.

1. Subtract 2 from both sides: 3x=18.
2. Divide by 3: x=6.
3. Check: 3\times6+2=20.

Undo subtracting 7 before dividing.

1. Add 7 to both sides: 5x=25.
2. Divide by 5: x=5.
3. Check: 5\times5-7=18.

After undoing the addition, undo division by 2.

1. Subtract 6 from both sides: \frac{x}{2}=5.
2. Multiply by 2: x=10.
3. Check: 10\div2+6=11.

A decimal multiplier is still undone by division.

1. Divide both sides by 2.5: x=15\div2.5.
2. x=6. Check: 2.5\times6=15.

Keep the sign of the result after subtracting; the answer need not be a whole number.

1. Subtract 9 from both sides: 4p=-6.
2. Divide by 4: p=-\frac64=-1.5.
3. Check: 4\times(-1.5)+9=-6+9=3.

Undo the subtraction, then the division.

1. Add 3 to both sides: \frac{x}{5}=7.
2. Multiply by 5: x=35.
3. Check: 35\div5-3=7-3=4.

Let one notebook cost n pounds. Write the total cost using three notebooks and one pen.

1. Let one notebook cost n pounds. The total cost gives 3n+2=17.
2. Subtract 2 from both sides: 3n=15.
3. Divide by 3: n=5. One notebook costs £5.
4. Check: 3\times5+2=17.

Work backwards in reverse order: undo the last operation first.

1. Let the original number be x. The equation is \frac{x}{4}-3=5.
2. Add 3 to get \frac{x}{4}=8, then multiply by 4 to get x=32.
3. Check: 32\div4-3=8-3=5.

The perimeter is the sum of all four sides, not the area.

1. The perimeter includes two lengths and two widths, so 2x+2\times5=34.
2. Simplify: 2x+10=34. Subtract 10 to get 2x=24.
3. Divide by 2: x=12 cm. Check: 12+5+12+5=34 cm.

For (a), ask which operation undoes +8. For (b), start again from the original equation.

Part (a)

1. The original left side has 8 added to 3x. To remove it, subtract 8 from both sides.
2. Jade added 8 to 26 instead; that does not undo the +8.

Part (b)

1. Subtract 8: 3x=26-8=18.
2. Divide by 3: x=6.
3. Check: 3\times6+8=26.

Add the three side expressions to form an equation for the perimeter.

1. Add the three sides: (x+1)+(x+4)+(2x-2)=31.
2. Collect like terms: 4x+3=31.
3. Subtract 3: 4x=28. Divide by 4: x=7.
4. The sides are 8, 11 and 12 cm. Their total is 31 cm.

After removing the 14, remember that the multiplier of x is negative.

1. Subtract 14 from both sides: -3x=15.
2. Divide both sides by -3: x=-5.
3. Check: 14-3\times(-5)=14+15=29.

For (a), remove the amount poured into the glass before sharing. For (b), convert litres to millilitres.

Part (a)

1. Let each jar hold j litres. The amounts give 4j+0.3=1.8.
2. Subtract 0.3: 4j=1.5. Divide by 4: j=0.375 litres.
3. Check: 4\times0.375+0.3=1.8 litres.

Part (b)

1. Multiply the answer in litres by 1000: 0.375\times1000=375.
2. Each jar contains 375 ml.

For (a), undo the fixed charge and then the per-mile multiplication. For (b), calculate a new total using the original charging rule.

Part (a)

1. Let the distance be d miles. Then 1.60d+3.80=13.40.
2. Subtract 3.80: 1.60d=9.60. Divide by 1.60: d=6 miles.

Part (b)

1. The fixed £3.80 is still charged once.
2. Use the forward rule: 3.80+1.60\times9=3.80+14.40=18.20.
3. The 9-mile fare is £18.20.