How do you solve #x^2+y^2=109# and #x-y= - 7# using substitution?
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#:."The Solution Set="{(x,y)=(3,10),(-10,-3)}#.
#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)}#.
#x = - 10 , y = -3 and x = 3, y = 10 #
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#
Express the implicit relation in terms of one variable only. See method below.
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)#
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)
:}#
