Given the simultaneous equations 2x - y = 4 and x^2 + y^2 = 13, where both x and y are pos… — JAMB UTME

Given the simultaneous equations $2x - y = 4$ and $x^2 + y^2 = 13$ , where both $x$ and $y$ are positive real numbers, calculate the value of $x + y$ .

Answer: 5

Answer

The correct value of

x + y

is 5.

Substituting

y = 2x - 4

into

x^2 + y^2 = 13

gives

5x^2 - 16x + 3 = 0

, which yields

x = 3

x = 0.2

. The corresponding

y

-values are

y = 2

and

y = -3.6

. Since both

x

and

y

must be positive, the valid pair is

(3, 2)

, giving

x + y = 3 + 2 = 5

Step-by-Step Solution

Express

y

in terms of

x

using the linear equation

2x - y = 4

y = 2x - 4

Substitution is the standard method for solving simultaneous linear and quadratic equations.

Substitute

y = 2x - 4

into the quadratic equation

x^2 + y^2 = 13

x^2 + (2x - 4)^2 = 13

This reduces the system to a single quadratic equation in one variable.

Expand and collect like terms.

5x^2 - 16x + 3 = 0

Expanding

(2x - 4)^2 = 4x^2 - 16x + 16

and subtracting 13 puts the quadratic in standard form

ax^2 + bx + c = 0

Solve the quadratic equation

5x^2 - 16x + 3 = 0

for

x

x = 3

x = 0.2

Factoring

(5x - 1)(x - 3) = 0

yields two real solutions for

x

Find corresponding values of

y

and apply the positivity constraint

x > 0

and

y > 0

x = 3, y = 2

When

x = 0.2

y = -3.6

, which is not positive. Hence,

(3, 2)

is the only valid solution pair.

Calculate

x + y

3 + 2 = 5

Key Concept

Solving simultaneous linear and quadratic equations using substitution

Estimated Time:1m 30s

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