Question

Solve algebraically the simultaneous equations


`x^(2)` +
`y^(2)` = 25

y = 2x -2

Give your final pairs of values in the form,

x = . . . , y = . . . and x = . . ., y = . . .,

Answer 1

`x = 3`
`y = 4`
`x = -(7)/(5)`
`y = -(24)/(5)`

Explanation:

y = 2x - 2

substitute y with 2x - 2 in the other equation

`x^(2) +y^(2) = 25`

`x^(2) +(2x-2)^(2) = 25`

`x^(2) + 4x^(2) -4x-4x+4 = 25`

`5x^(2) -8x+4=25`

`5x^(2) -8x-21=0`

Use the quadratic equation to solve

`\frac{-b(+-)\sqrt{b^(2)-4ac }}{2a}`

`a = 5\\b = -8 \\c = -21`

`\frac{-(-8)(+-)\sqrt{(-8)^(2)-4(5)(-21) }}{2(5)}`

`(8(+-)√(64-(-420) ))/(10)`

`(8+√(484 ))/(10)`
`(8-√(484 ))/(10)`

`x` =
`3`
`x` =
`-(7)/(5)`

Now use these values of x in the other equation to find two values of y

y = 2x - 2 y = 2x - 2

y = 2(3) - 2 y = 2(-7/5) - 2

y = 4 y = - (24/5)