Algebra, Year 10: simultaneous equations and iteration

x = 7 and y = 3

HintOne equation has +y and the other has −y. Think what happens if the two equations are combined.

The whyAdding the equations makes the y terms cancel: 2x = 14, so x = 7. Putting x = 7 into x + y = 10 gives y = 3. Check in the other equation: 7 − 3 = 4.

12; 4; 2

HintDoubling makes the y terms match, so that taking one equation from the other removes them.

The whyThe y terms must match before they can be eliminated. After doubling, (3x + 2y) − (2x + 2y) = x and 16 − 12 = 4. Then x + y = 6 gives y = 2. Check in the first equation: 3 × 4 + 2 × 2 = 16.

x = 5 and y = 15

HintThe first equation already says what y is worth in terms of x. Use that in the second equation.

The whyReplacing y by 3x in the second equation gives x + 3x = 20, so 4x = 20 and x = 5. Then y = 3 × 5 = 15. This method is called substitution, and it is quickest when one equation already has a letter as its subject.

A coffee costs £2.50 and a tea costs £2.00

HintCompare the two orders: what is in the first one that is not in the second?

The whyWith c for a coffee and t for a tea: 2c + t = 7 and c + t = 4.5. Subtracting the second from the first leaves one coffee: c = 2.50. Then 2.50 + t = 4.50, so t = 2.00. Check: 2 × 2.50 + 2.00 = 7.00.

Elimination

HintThe word means getting rid of something, as when a team is knocked out of a cup.

The whyIf the two equations have the same number of x or of y, adding or subtracting them leaves an equation with only one unknown. If they do not match, multiply one or both equations first so that they do.

Iteration

HintThe word comes from the Latin for "again".

The whySome equations cannot be solved exactly by rearranging. Instead a formula such as xₙ₊₁ = (xₙ + 6)/2 is applied to a starting value, then to the result, and so on, until the values settle down to the solution.

4; 5; 5.5

HintEach new value is made from the one just before it, not from the starting value.

The whyx₁ = (2 + 6)/2, x₂ = (4 + 6)/2 and x₃ = (5 + 6)/2. The small numbers are labels that count the steps; they are not powers or multipliers. The values are closing in on 6, the solution of x = (x + 6)/2.

179

HintWork out the first year, then use that result as the start of the second year.

The whyP₁ = 1.1 × 200 − 30 = 190. Then P₂ = 1.1 × 190 − 30 = 209 − 30 = 179. The same rule is applied again to each new answer, which is what makes it an iterative process.

It has a solution between x = 1 and x = 2

HintOne output is negative and the other positive. Think about what the graph must do on the way from one to the other.

The whyThe graph of y = x³ + x − 5 is below the x-axis at x = 1 and above it at x = 2. It is a continuous curve, so it must cross the axis somewhere in between, and a crossing is a solution. Iteration can then pin the value down (it is about 1.52).

When two values in a row agree once they are rounded to 2 decimal places

HintWatch how much each new value differs from the one before.

The whyAs an iteration closes in on a solution, the values change less and less. Once the change is too small to alter the second decimal place, further steps will not change the rounded answer. Keep all the calculator digits while iterating and round only at the end.

Add 3x + 1 to both sides to get x³ = 3x + 1, then take the cube root of both sides

HintAim to leave only the cubed term on the left.

The whyx³ − 3x − 1 = 0 becomes x³ = 3x + 1, and taking the cube root gives x = ∛(3x + 1). An x on each side is what an iteration needs: the formula becomes xₙ₊₁ = ∛(3xₙ + 1), with the old value on the right and the new one on the left.