How do you use substitution to solve the pair of equations: #y=2x-3# and #x^2+y^2=16#?

1 Answer
Dec 6, 2016

Answer:

#(x,y)~~color(green)(""(2.885,2770)) or color(green)(""(-0.485,-3.970))#

Explanation:

Given
[1]#color(white)("XXX")y=2x-3#
[2]#color(white)("XXX")x^2+y^2=16#
Using [1] we know that we can substitute #(2x-3)# for #y# in [2]
[3]#color(white)("XXX")x^2+(2x-3)^2=16#

[4]#color(white)("XXX")rarr x^2+4x^2-12x+9=16#

[5]#color(white)("XXX")rarr5x^2-12x-7=0#

Using the quadratic formula:
[6]#color(white)("XXX")x=(12+-sqrt((-12)^2-4 * 5 * (-7)))/(2 * 5)#

simplifying (and using a calculator)
#color(white)("XXX"){:(x~~2.885," or ",x~~-0.485):}#
using [1] #rarr#
#color(white)("XXX"){:(y~~2.770," or ", y~~-3.970):}#

The image of the given relations (below) indicates that these results are reasonable.
enter image source here