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

2 Answers
Apr 29, 2017

-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

Apr 29, 2017

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 "

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