If a and b are real numbers satisfying the simultaneous equations a + b = 10 and a^2 - b^2… — JAMB UTME

If $a$ and $b$ are real numbers satisfying the simultaneous equations $a + b = 10$ and $a^2 - b^2 = 40$ , what is the value of $a$ ?

Answer: 7

The value of

a

is 7.

Using the difference of squares identity,

a^2 - b^2 = (a+b)(a-b)

. Substituting

a+b = 10

into

a^2 - b^2 = 40

gives

10(a-b) = 40

, which simplifies to

a-b = 4

. Adding the two linear equations

a+b = 10

and

a-b = 4

eliminates

b

, giving

2a = 14

, hence

a = 7

Factorize the quadratic expression

a^2 - b^2

(a + b)(a - b) = 40

Apply the difference of two squares identity.

Substitute

a + b = 10

into the factorized equation.

10(a - b) = 40 \implies a - b = 4

Simplifying yields a second linear equation.

Solve the system of linear equations

a + b = 10

and

a - b = 4

for

a

(a + b) + (a - b) = 10 + 4 \implies 2a = 14 \implies a = 7

Adding the two equations eliminates

b

directly.

Simultaneous linear and quadratic equations involving difference of squares

Question