Guide7 patterns

Algebra

Solving linear equations and simultaneous equations.

Linear equation

The method

Undo the constant first (add or subtract it), then divide by the coefficient.

Worked example

Solve for x: 5x − 7 = 28

Answer: 7

Add 7: 5x = 35. Divide by 5: x = 7. The 4.2 foil divides 28 by 5 and ignores the −7; 35 forgets to divide; 21 just computes 28−7.

Substitution

The method

Substitute the value carefully, squaring only what's squared, then evaluate term by term.

Worked example

If y = 3x² − 4x + 2, what is y when x = 5?

Answer: 57

Substitute x=5: 3×25 − 4×5 + 2 = 75 − 20 + 2 = 57. The 207 foil squares (3×5) instead of just x; 97 adds the −4x term; 53 subtracts the +2.

Rearranging formulas

The method

Rearrange to isolate the unknown before substituting the numbers.

Worked example

Given v = u + at, make a the subject and find a when v = 30, u = 6, t = 4.

Answer: 6

Rearrange: a = (v − u)/t = (30 − 6)/4 = 24/4 = 6. The 9 foil adds u instead of subtracting; 24 forgets to divide by t.

Simultaneous equations

The method

Add or subtract the equations to eliminate one unknown, solve, then back-substitute.

Worked example

Solve the pair and give x: 2x + y = 11 and x − y = 1

Answer: 4

Add the equations to eliminate y: 3x = 12, so x = 4 (and y = 3). The 3 foil gives y; 12 forgets to divide by 3.

Inequalities

The method

Solve the inequality, then pick the smallest or largest whole number that satisfies it strictly.

Worked example

For the smallest whole number n, 3n + 4 is greater than 25. What is n?

Answer: 8

3n + 4 > 25 means 3n > 21, so n > 7. The smallest whole number greater than 7 is 8. The 7 foil includes the boundary (3·7+4 = 25, which is not greater than 25).

Word problems

The method

Translate the sentence into an equation (total = fixed + rate × quantity), then solve step by step.

Worked example

A technician charges a £45 call-out fee plus £30 for each hour worked. A repair bill came to £195. How many hours did the technician work?

Answer: 5

Translate the words: total = fee + rate × hours, so 195 = 45 + 30h. Subtract the fee: 30h = 150. Divide by the rate: h = 5 hours. The 6.5 foil divides the whole bill by £30 and forgets the £45 fee. The 2.6 foil divides by fee and rate added together. The 5.5 foil subtracts one hour's rate (£30) instead of the £45 fee.

Age problems

The method

Set the unknown as x, express the other quantities in terms of x, build the equation from the condition, and solve.

Worked example

A father is four times as old as his son. In five years, he will be three times as old. How old is the son now?

Answer: 10

Let the son be x, the father 4x. In five years: 4x + 5 = 3(x + 5) → 4x + 5 = 3x + 15 → x = 10. The 40 foil gives the father's age now; 45 the father in five years; 15 the son in five years.