Question

How do you solve `-x + y = 1` and `x^2 +y^2 = 16` using substitution?

Answer 1

`-x +y = 1`
`y = x+1` -----(1)

`x^2 + y^2 = 16` -----(2)

Substituting (1) into (2):
`x^2 + (x+1)^2 = 16`
`x^2 + x^2 + 2x + 1 = 16`
`2 x^2 + 2x - 15 = 0`

Solving the quadratic equation for x,
`x= (sqrt31 -1) /2 or (-sqrt31 -1) /2`

Substituting x into (1) to find y,
`y = (sqrt31 +1) /2 or (-sqrt31 +1) /2`

Answer 2

Bit long as I have explained a lot of the steps. See qin's solution for a shorter version (jumped steps)

`(2.28,3.28) and (-3.28,-2.28)` to 2 dp

Explanation:

Recognise that `x^2+y^2=4^2` is a Pythagorean (Pythagoras) representation of a circle

`x^2+y^2=4^2" "..................Equation(1)`
`-x+y=1" "...................Equation(2)`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(magenta)("Consider "Equation(2))`

Add `x` to both sides giving:

`y=x+1" "...............Equation(2_a)`

Using `Equation(2_a)` substitute for `y` in `Equation(1)`

`color(green)(x^2color(red)(+y^2)=4^2" "->" "x^2color(red)(+(x+1)^2)=16)" "..Equation(1_a)`
.............................................................................................
`color(magenta)("Consider "(x+1)^2)`

But `(x+1)^2->color(red)((x+1))color(blue)((x+1))`

Multiply everything inside the right bracket by everything inside the left bracket.

`color(red)(xcolor(blue)((x+1))" "+" "1color(blue)((x+1))`
`color(white)(.)x^2+x" "color(white)(.)+" "x+1`

`x^2+2x+1`
.....................................................................................
`color(magenta)("Putting it all together to determine "x)`

`Equation(1_a)` becomes:

`x^2+x^2+2x+1=16`

`2x^2+2x+1=16" "->" "2x^2+2x-15=0`

Using the standardised form `y=ax^2+bx+c`

where `x=(-b+-sqrt(b^2-4ac))/(2a)`

`a=2"; "b=2"; "c=-15`

`=>x=(-2+-sqrt((2)^2-4(2)(-15)))/(2(2))`

`x=-1/2+-sqrt(124)/4`

`x=-1/2+-sqrt(2^2xx31)/4" "->" "x=-1/2+-1/2sqrt(31)`

`x~~2.28388... and x~~-3.28388...`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(magenta)("Determine "y" by substitution for "x" in "Equation(2_a))`

`y=x+1" "..............Equation(2_a)`

Set `x~~2.28388...`

`y~~3.28288...`

`ul(bar(|color(white)(2/2)"point "P_1->(x,y)=(2.28,3.28)" to 2 decimal places ")|)`

......................................................................................

Set `x~~-3.28388...`

`y~~-2.28288...`

`ul(bar(|color(white)(2/2)"point "P_2->(x,y)=(-3.28,-2.28)" to 2 decimal places "`