Question

If x = 1 and y = 2 is a solution to the simultaneous equation ax + by = 2 and bx + a^(2)y = 10, find the possible values of a and b.

Answer 1

This is the solution

Answer 2

Answer:

a = - 2, $\frac{9}{4}$ , b = 2, - $\frac{1}{8}$

Step-by-step explanation:

Since x = 1, y = 2 is a solution to the equations , then substitute these values into the 2 equations and solve for a and b

a + 2b = 2 → (1)

b + 2a² = 10 → (2)

In (1) subtract 2b from both sides

a = 2 - 2b → (3)

Substitute a = 2 - 2b into (2)

b + 2(2 - 2b)² = 10 ← expand parenthesis using FOIL

b + 2(4 - 8b + 4b²) = 10 ( simplify left side )

b + 8 - 16b + 8b² = 10 ( subtract 10 from both sides )

8b² - 15b - 2 = 0

Consider the factors of the product of the coefficient of the b² term and the constant term which sum to give the coefficient of the b- term

product = 8 × - 2 = - 16 and sum = - 15

The factors are - 16 and + 1

Use these factors to split the b- term

8b² - 16b + b - 2 = 0 ( factor first/second and third/fourth terms )

8b(b - 2) + 1(b - 2) = 0 ← factor out (b - 2) from each term

(b - 2)(8b + 1) = 0

Equate each factor to zero and solve for b

b - 2 = 0 ⇒ b = 2

8b + 1 = 0 ⇒ 8b = - 1 ⇒ b = - $\frac{1}{8}$

Substitute these values into (3) and evaluate for a

b = 2 ⇒ a = 2 - 2(2) = 2 - 4 = - 2

b = - $\frac{1}{8}$ ⇒ a = 2 - 2(- $\frac{1}{8}$ ) = 2 + $\frac{1}{4}$ = 2 $\frac{1}{4}$ = $\frac{9}{4}$