how to solve system of equation

Question

x^2+y=1

5x-y=-7

Answer 1

x² + y = 1
5x - y = -7

 

Don't let that power throw you off. We can still solve this like a linear pair. I'm going to suggest elimination in this case:

x² + y = 1
5x - y = -7

----------------

x² + 5x = -6

x² + 5x + 6 = 0

 

As you can see from here, we now have a quadratic formula.

Factoring it out, we get:

(x + 2)(x + 3) = 0

x = -2, -3

 

Let's put these values into our equations and see what happens:

5 * -2 - y = -7

10 - y = -7

-y = 3

y = -3

Check:

(-2)² + -3 = 1

4 - 3 = 1

1 = 1

 

5 * -3 - y = -7

-15 - y = -7

-y = 8

y = -8

Check:

(-3)² + -8 = 1

9 - 8 = 1

1 = 1

 

So the solutions are (-2, -3) and (-3, -8)

Answer 2

There are a number of ways to do this...

One nice trick is to eliminate a variable and solve a new single equation in a single variable.

 

For example in this problem try adding your equations together!

This should yield a quadratic equation in x, of the form ax^2+bx+c=0 for some constants a,b,c.

Apply the QuadraticFormula (factor and find solutions for x).  Plug back in to either original equation to find y.

 

Note there could be multiple solutions (or no solutions if we require x and y to be real numbers).