You may use substitution method, hence, you may write `x` in terms of `y,` in equation `2x + 2y = 7` , such that:

`2x + 2y = 7 => 2x = 7 - 2y => x = 7/2 - y`

Replacing `7/2 - y` for `x` in the bottom equation, yields:

`( 7/2 - y)^2 - 4y^2 = 8`

Expanding the square, yields:

`49/4 - 7y + y^2 - 4y^2 - 8 = 0`

You need to re-arrange the terms in descending powers of y, such that:

`-3y^2 - 7y + 17/4 = 0 => -12y^2 - 28y + 17 = 0 => 12y^2+ 28y - 17 = 0`

Using quadratic formula, yields:

`y_(1,2) = (-28+-sqrt(28^2 + 816))/24`

`y_(1,2) = (-28+-40)/24`

`y_1 = 12/24 => y_1 = 1/2`

`y_2 = 68/24 = y_2 = 17/6`

You need to evaluate `x_(1,2)` , such that:

`x_1 = 7/2 - y_1 =>x_1 = 7/2 - 1/2 => x_1 = 3`

`x_2 = 7/2 - y_2 => x_2 = 7/2 - 17/6 => x_2 = 4/6 => x_2 = 2/3`

**Hence, evaluating the solutions to the given system of equations, yields `x_1 = 3, y_2 = 1/2` and **`x_2 = 2/3, y_2 = 17/6.`

The set of equations 2x + 2y = 7 and x^2 - 4y^2 = 8 has to be solved.

2x + 2y = 7

=> `x = (7 - 2y)/2 = 3.5 - y`

Substitute in `x^2 - 4y^2 = 8`

=> `(3.5 - y)^2 - 4y^2 = 8`

=> `12.25 + y^2 - 7y - 4y^2 = 8`

=> `-3y^2 - 7y + 4.25 = 0`

`y = (7+-sqrt(49 + 51))/(-6)`

= `(7+- 10)/(-6)`

y = `-17/6` and y = `1/2`

x = 3.5 - y

For y = `-17/6` , x = `19/3` and for y = `1/2` , x = 3

**The solution of the given system of equations is `{3, 1/2}` and **`{19/3, -17/6}`

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