Question

How do you solve this system of equations: `5x ^ { 2} - 2y = 1 and 4x + 5y = 47`?

Answer 1

Solution is `(-2.1564,11.125)` and `(1.8364,7.931)`

Explanation:

While the first equation `5x^2-2y=1` represents a parabola, `4x+5y=47` represents a straight line. Hence, there is a possibility of up to two solutions.

To solve them, wecan use substitution method. Drom first equation we get `2y=5x^2-1` and hence `5y=(5x^2-1)xx5/2=25/2x^2-5/2` and putting this in the equation for straight line, we get

`4x+25/2x^2-5/2=47`

or `25x^2+8x-99=0`

and `x=(-8+-sqrt(8^2-4xx25xx(-99)))/50`

= `(-8+-sqrt(64+9900))/50`

= `(-8+-99.82)/50` i.e. `-2.1564` or `1.8364`

For `x=-2.1564`, `y=(47-4xx(-2.1564))/5=11.125`

and for `x=1.8364`, `y=(47-4xx1.8364)/5=7.931`

Hence, the solution is `(-2.1564,11.125)` and `(1.8364,7.931)`

graph{(5x^2-2y-1)(4x+5y-47)=0 [-10.75, 9.25, 2.2, 12.2]}