X²+4y=5 And y-2x=5 using substitution method to solve it

Answer 1

I'm afraid Patti's answer is incorrect.

Her methodology is good, but her arithmetic has a mistake.

x² + 4 (2x +5) = 5

x² + 8x+ 20 = 5

x² + 8x+15 = 0

(x+5)(x+3)=0

x= -3, -5

Answer 2

To use substitution, you need to solve one equation for one of the variables.

In the second equation, solve for y by adding 2x to both sides.

y = 2x + 5, now substitute 2x + 5 for every y in the second equation.

x² + 4 (2x +5) = 5 the bold print is the substitution. Now so√√lve the quadratic

x² + 8x + 10 = 5

x² + 8x + 5 = 0

x =( -8 +/- √(64 - 20))/ 2

x = (-8 +/- √ 44)/2

x= -4 +/- √11