Question

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

Answer 1

`:."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)}`.

Answer 2

`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`

Answer 3

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`

Answer 4

`(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) :}#