If x and y are real numbers satisfying the simultaneous equations x - y = 4 and x^2 + y^2 … — JAMB UTME

If $x$ and $y$ are real numbers satisfying the simultaneous equations $x - y = 4$ and $x^2 + y^2 = 26$ , what is the value of the product $xy$ ?

Answer: 5

Answer

The value of the product

xy

is 5.

Expressing

x

from the linear equation yields

x = y + 4

. Substituting this into

x^2 + y^2 = 26

gives

(y + 4)^2 + y^2 = 26

, which expands and simplifies to

y^2 + 4y - 5 = 0

. Factoring gives solutions

y = 1

(with

x = 5

) and

y = -5

(with

x = -1

). Both solution pairs

(5, 1)

and

(-1, -5)

result in the product

xy = 5

Step-by-Step Solution

Express

x

in terms of

y

using the linear equation

x = y + 4

Isolate

x

to substitute into the second equation.

Substitute

x = y + 4

into the quadratic equation

x^2 + y^2 = 26

(y + 4)^2 + y^2 = 26

Reduce the system to a single quadratic equation in

y

Expand and simplify into standard quadratic form

y^2 + 4y - 5 = 0

Expanding gives

2y^2 + 8y + 16 = 26

, which simplifies by subtracting 26 and dividing by 2.

Solve for

y

by factoring

y = 1

y = -5

The factors of

y^2 + 4y - 5

are

(y + 5)(y - 1) = 0

Compute corresponding

x

values and the product

xy

For

y = 1

x = 5 \implies xy = 5

; for

y = -5

x = -1 \implies xy = 5

Substitute each

y

back into

x = y + 4

and evaluate

xy

Key Concept

Simultaneous Linear and Quadratic Equations

Question

Answer

Step-by-Step Solution

Key Concept

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