Question 110604
using the two equations below, solve for x and y using three different methods, substitution, addition and graphing.
-2x + 3y = 6
 3x - 5y =-11
:
Substitution:
Rearrange the 1st equation to be used for subsitution:
3y = 2x + 6
y = {{{2/3}}}x + {{{6/3}}}; divided equation by 3
y = {{{2/3}}}x + 2
:
Substitute ({{{2/3}}}x + 2) for y in the 2nd equation:
3x - 5({{{2/3}}}x + 2) = -11
:
3x - {{{10/3}}}x - 10 = -11
;
{{{9/3}}}x - {{{10/3}}}x = -11 + 10
:
-{{{1/3}}}x = -1
:
{{{1/3}}}x = 1; multiplied by -1 to get rid of the negatives
:
x = 3; Multiplied by 3 to get rid of the fraction
:
Find y using the y = {{{2/3}}}x + 2
y = {{{2/3}}}3 + 2
y = 2 + 2
y = 4
:
Our solution x = 3; y = 4
:
Substitute for x and y in the 2nd equation to confirm our solutions:
3x - 5y =-11
:
3(3) - 5(4) =
9 - 20 = -11
:
:
Elimination:
-2x + 3y = 6
 3x - 5y =-11
:
Multiply the 1st equation by 3 and the 2nd equation by 2:
-6x + 9y = 18
+6x -10y = -22
---------------adding eliminates x, find y:
0x - 1y = -4
y = +4
Find x just like we did before using substitution for y in either equation
:
To graph this:
y = {{{2/3}}}x + 2; from above:
and
 3x - 5y =-11
-5y = - 3x - 11
5y = 3x + 11; multiplied by -1 to get rid of all those negatives
y = {{{3/5}}}x + {{{11/5}}}
:
Plot these to equations, should look like this
{{{ graph( 500, 500, -6, 8, -2, 5, (2/3)x +2, .6x+(11/5)) }}}
:
Not very clear but they should intersect at x=3, y=4
:
How about this? Did it explain these three methods to you?