This problem just "begs" to be done by elimination, look at the y values
add #1 and #2 2x + 3z = 14 #4
add #2 and #3 3x + 5z = 22 #5
triple #4 -- 6x + 9z = 42
double #5 -- 6x + 10z = 44
subtract #5 - #4
z = 2
sub into #4 -- 2x+6 = 14 , x = 4
back into #1
4 + y + 2 = 7
y = 1
I did a problem and used the substitution method and it took me forever to solve by hand. Is there a shorter way to show how this problem was done. The problem was
x+y+z=7,x-y+2z=7,2x+y+3z=15
The final answer came out to be
x=4
z=2
y=1
1 answer
1 answer