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

Answer:

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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