We have three equations:
#x + 2y + z = 4# (Equation 1.1)
#3x + 5y - 7z = 1# (Equation 1.2)
#2x + 7y - z = 8# (Equation 1.3)
Let's try to solve for one of the variables using elimination. This means that we're going to have to cancel out the two other variables by manipulating the equations and adding. Let's make the coefficient of #z# become #-7# in all equations by multiplying each equation accordingly:
#(x + 2y + z) * (-7) = 4 * (-7)#
#(3x + 5y - 7z) * 1 = 1 * 1#
#(2x + 7y - z) * 7 = 8 * 7#
Let's distribute:
#-7x - 14y - 7z = -28# (Equation 2.1)
#3x + 5y - 7z = 1# (Equation 2.2)
#14x + 49y - 7z = 56# (Equation 2.3)
Now, let's take any two equations... perhaps Equation 2.1 and Equation 2.2, and subtract one from the other:
#(-7x - 14y - 7z) - (3x + 5y - 7z) = (-28) - (1)#
#-7x - 14y - 7z - 3x - 5y + 7z = -29#
#-10x - 19y = -29# (Equation 3.1)
Notice, the variable #z# has canceled out! However, we need another of these equations to finally solve for one variable. Let's subtract Equation 2.2 from Equation 2.3 now:
#(14x + 49y - 7z) - (3x + 5y - 7z) = (56) - (1)#
#14x + 49y - 7z - 3x - 5y + 7z = 55#
#11x + 44y = 55# (Equation 3.2)
Alright! So we have Equation 3.1 and Equation 3.2, as follows:
#-10x - 19y = -29#
#11x + 44y = 55#
We're going to do the same thing we did to Equation 1.1, Equation 1.2 and Equation 1.3: try to make one of the coefficents the same! Let's try to make the coefficient of #x# become #110#:
#(-10x - 19y) * (-11) = (-29) * (-11)#
#(11x + 44y) * (10) = (55) * (10)#
And multiply them over:
#110x + 209y = 319# (Equation 4.1)
#110x + 440y = 550# (Equation 4.2)
Subtract Equation 4.1 from Equation 4.2:
#(110x + 440y) - (110x + 209y) = (550) - (319)#
#110x + 440y - 110x - 209y = 231#
#231y = 231#
Sweet! Divide by #231#:
#(231y)/231 = 231/231#
#y = 1#
Alright... substitute that into Equation 1.1, Equation 1.2 and Equation 1.3:
#x + 2(1) + z = 4#
#3x + 5(1) - 7z = 1#
#2x + 7(1) - z = 8#
We could subtract each constant, "moving" them over to the right:
#x + z = 2# (Equation 5.1)
#3x - 7z = -4# (Equation 5.2)
#2x - z = 1# (Equation 5.3)
Interesting! We can simply add Equation 5.3 to Equation 5.1:
#(x + z) + (2x - z) = (2) + (1)#
#3x = 3#
#x = 1#
Huh. Now we just need to solve for #z#. We can substitute this result to Equation 5.1:
#(1) + z = 2#
#z = 1#
That's unexpected! So we have #x = 1#, #y = 1#, and #z = 1#. Let's substitute these into Equation 1.1, Equation 1.2, and Equation 1.3 to verify:
#(1) + 2(1) + (1) = 4#
#3(1) + 5(1) - 7(1) = 1#
#2(1) + 7(1) - (1) = 8#
Yep! These solutions can be rewritten as #(1, 1, 1)#, representing #(x, y, z)#.
