To form a system from two constraints, name the unknowns, write each condition as one equation, then solve the pair by substitution. One condition is often linear, such as a sum or a difference, and the other is nonlinear, such as a product, an area or Pythagoras’ theorem.
This lesson uses the elimination skills from substituting a line into a circle and eliminating a variable from a nonlinear pair, and it sits in simultaneous linear and nonlinear models.
How do you turn words into equations?
Read the question once to find what is unknown, then once more to find the facts. Each fact becomes an equation.
| Wording | Equation |
|---|---|
| perimeter of a rectangle is P | 2x + 2y = P, so x + y = P/2 |
| area is A | xy = A |
| diagonal is d | x² + y² = d² |
| two numbers differ by 3 | x − y = 3 |
| sum of squares is S | x² + y² = S |
Then choose the linear equation, rearrange it, and substitute into the nonlinear one. A linear equation keeps the working short.
Worked example
A rectangular garden has a perimeter of 34 m and a diagonal of 13 m. Find its length and width.
Define: let the length be x m and the width y m.
Condition 1, perimeter: 2x + 2y = 34, so x + y = 17.
Condition 2, diagonal: by Pythagoras, x² + y² = 13² = 169.
Substitute: y = 17 − x, so x² + (17 − x)² = 169.
Expand: x² + 289 − 34x + x² = 169, so 2x² − 34x + 120 = 0, then x² − 17x + 60 = 0.
Solve: (x − 5)(x − 12) = 0, so x = 5 or x = 12.
Partners: if x = 5, y = 12, and if x = 12, y = 5. These are the same rectangle with the labels swapped.
Answer: the garden is 12 m by 5 m.
Check: 2(12 + 5) = 34, and 12² + 5² = 144 + 25 = 169.
The mistake to watch for
The typical error is to use the whole perimeter as if it were x + y.
Mistaken working: perimeter 34 gives x + y = 34.
The student forgot that a rectangle has two lengths and two widths, so x + y is half the perimeter.
With x + y = 34 the substitution gives x² + (34 − x)² = 169, which leads to 2x² − 68x + 987 = 0. Its discriminant is 68² − 4(2)(987) = 4624 − 7896, which is negative, so there are no real solutions.
A result that says the rectangle cannot exist is a signal to return to the first equation. The correction is to halve the perimeter: a rectangle’s perimeter is 2x + 2y, so x + y = 17.
Check yourself
Work these on paper, then open each answer.
1. Two numbers add to 10 and multiply to 21. Find them.
Show answer
x + y = 10 and xy = 21. Substitute y = 10 − x: x(10 − x) = 21, so x² − 10x + 21 = 0 and (x − 3)(x − 7) = 0.
The numbers are 3 and 7. Check: 3 + 7 = 10 and 3 × 7 = 21.
2. A right-angled triangle has a hypotenuse of 13 cm, and its shorter leg is 7 cm shorter than the longer leg. Find both legs.
Show answer
Let the legs be x (longer) and y (shorter). Then x − y = 7 and x² + y² = 169. Substitute x = y + 7: (y + 7)² + y² = 169, so 2y² + 14y + 49 = 169 and y² + 7y − 60 = 0. Then (y + 12)(y − 5) = 0, so y = −12 or y = 5.
A length cannot be negative, so reject y = −12. Then y = 5 and x = 12.
The legs are 12 cm and 5 cm. Check: 144 + 25 = 169.
3. A rectangle has area 48 cm² and perimeter 28 cm. Find its sides.
Show answer
xy = 48 and x + y = 14. Substitute y = 14 − x: x(14 − x) = 48, so x² − 14x + 48 = 0 and (x − 6)(x − 8) = 0.
The sides are 6 cm and 8 cm. Check: 6 × 8 = 48 and 2(6 + 8) = 28.
Where this leads next
Once you can form the system reliably, the next step is asking how many solutions it has, in checking parameter values for a requested number of intersections. The non-calculator working trainer keeps exact arithmetic tidy, and the quadratic structure explorer shows what the resulting quadratic looks like.
If translating wording into equations is where your marks go, our teachers can practise that with fresh contexts in online one-to-one Additional Mathematics tuition.