Question

`\left \{ {{x^2+y^2=34} \atop {x+3y=18}} \right.`
Solve

Answer 1

see explanation

Explanation:

given the 2 equations

x² + y² = 34 → (1)

x + 3y = 18 → (2)

rearrange (2) by subtracting 3y from both sides

x = 18 - 3y → (3)

Substitute x = 18 - 3y into (1)

(18 - 3y)² + y² = 34

324 - 108y + 9y² + y² = 34

10y² - 108y + 324 = 34 ( subtract 34 from both sides )

10y² - 108y + 290 = 0 ← in standard form

divide through by 2

5y² - 54y + 145 = 0

(5y - 29)(y - 5) = 0 ← in factored form

Equate each factor to zero and solve for y

y - 5 = 0 ⇒ y = 5

5y - 29 = 0 ⇒ y =
`(29)/(5)`

Substitute these values into (3) for corresponding values of x

y = 5 : x = 18 - 15 = 3 ⇒ (3, 5) ← is a solution

y =
`(29)/(5)` : x = 18 -
`(87)/(5)` =
`(3)/(5)`

⇒ (
`(3)/(5)`,
`(29)/(5)` ) ← is a solution

Answer 2

it doesn't need to solve it hardly. just try some natural numbers for the first one and check it in the second. so you can conclude they would be 5 , 3.