How do you solve #x^2+y^2=109# and #x-y= - 7# using substitution?

4 Answers
Mar 28, 2018

#:."The Solution Set="{(x,y)=(3,10),(-10,-3)}#.

Explanation:

#x-y=-7 rArr x=y-7#.

Sub.ing in, # x^2+y^2=109#, we get,

# (y-7)^2+y^2=109, or, #

# 2y^2-14y+49-109=0, i.e., #

# y^2-7y-30=0#.

Using # 10xx3=30, and, 10-3=7#, we get,

# ul(y^2-10y)+ul(3y-30)=0#.

#:. y(y-10)+3(y-10)=0#.

#:. (y-10)(y+3)=0#.

#:. y=10, or, y=-3#.

# y=10, &, x=y-7 rArr x=3,and, y=-3 rArr x=-10#.

These roots satisfy the given eqns.

#:."The Solution Set="{(x,y)=(3,10),(-10,-3)}#.

Mar 28, 2018

#x = - 10 , y = -3 and x = 3, y = 10 #

Explanation:

Rearrange #x - y = -7# so #x = y - 7#
Now substitute "#y - 7# " for the #x# in the first equation.

#x^2 + y^2 = 109# becomes #(y - 7)^2 + y^2 = 109#
Expand the bracket:
#y^2 - 7y - 7y +49 + y^2 = 109#

Collect like terms:
#2y^2 - 14y + 49 = 109#

subtract 109 from both sides:
#2y^2 - 14y - 60 = 0#

Factorise:
#(2y + 6)(y - 10) = 0#

so #2y + 6 = 0# or #y - 10 = 0#

then #y = -3# or #y = 10#

Now put these two values into #x - y = - 7#
when #y = - 3#,
#x - (- 3) =- 7#
#x + 3 = - 7#
#x = - 10#

when #y = 10#,
#x -10 = - 7#
#x = 3#

So the answers are:
#x = - 10 and y = - 3#
and
#x = 3 and y = 10#

Mar 28, 2018

Express the implicit relation in terms of one variable only. See method below.

Explanation:

You have two equations:

#x^2 + y^2 = 109#

#x-y=-7#

You can express #x# in terms of #y# or express #y# in terms of #x#. Either way is fine.

Since #x-y=-7#, that means #y=x+7#

Using this substitution we can write #x^2 + (x+7)^2 = 109#

This will give us a quadratic expression where we can solve for #x#.

#x^2 + (x+7)^2 = x^2 + (x^2 + 14x + 49) = 109#

Simplifying this we get:
#x^2 + 7x -30 = 0#

This can be factorized to #(x-3)(x+10)=0# giving us the solutions #x = 3# and #x=-10#.

We can then solve for #y# using the equation #y=x+7#

If #x=3# then #y=10#

If #x=-10# then #y=-3#

#(x,y)=(3,10)color(white)("xxx")"or"color(white)("xxx")(x,y)=(-10,-3)#

Explanation:

Given
[1]#color(white)("XXX")x^2+y^2=109#
[2]#color(white)("XXX")x-y=-7#

Note that [2] implies
[3]#color(white)("XXX")x=y-7#

Substituting #(y-7)# for #x# in [1]
[4]#color(white)("XXX")(y-7)^2+y^2=109#

Expanding and simplifying the left side of [4]
[5]#color(white)("XXX")2y^2-14y+49=109#

Converting to standard polynomial form by subtracting #109# from both sides
[6]#color(white)("XXX")2y^2-14y-60=0#

Factoring
[7]#color(white)("XXX")2(y-10)(y+3)=0#

Which implies:
#{: ([8a]color(white)("XX")y=10,color(white)("XX")"or"color(white)("XX"), [8b]color(white)("XX")y=-3), ("Substituting "10" for "x,,"Substituting "-3" for "x), ("in [3]",,"in [3]"), ([9a]color(white)("XX")x=10-7=3,,[9b]color(white)("XX")x=-3-7=-10) :}#