Re: x + 2y + 3z = 4 and 5x + 4y + 3z = 8. What is the value of x + y + z?
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19 Mar 2023, 18:29
A few ways to solve this:
1. Elimination:
\(x+2y+3z=4\\
5x+4y+3z=8\)
eliminate z
\(4x+2y=4\\
=2x+y=2\\
=y=2-2x\)
not very useful by itself, so next we try to eliminate y
\(5x++4y+3z=8\\
2x+4y+6z=8\)
=
\(3x-6z=0\\
3x=6z\\
x=2z\)
useful, let's plug this in to the original equation and see what happens
now we can replace the x in the original equation
\(x+2y+3z=4\\
(2z)+2y+3z=4\\
2y+5z=4\)
now since \(y=2-2x=2-2(2z)=2-4z\)
replace y
\(2(2-4z)+5z=4\\
4-8z+5z=4\\
-3z=0\\
z=0\)
and since
\(x=2z\\
x=0\)
if both x and z = 0
\(x+2y+3z=4\\
0+2y+0=4\\
2y=4\\
y=2\)
so \(x+y+z=0+2+0=2\)
method 2: try to find the entire equation without eliminating down to 1 variable (might be faster)
We have the equation
\(x+2y+3z=4\\
5x+4y+3z=8\)
the equation lines up nicely that there is a value 3z, let's work on that and see where it takes us
so
\(5x+4y+3z=8\\
minus\\
x+2y+3z=4\\
=\\
4x+2y=4\\
2x+y=2\\
\)
we need to find some version of y+2z to add them together to 2x+2y+2z=something
lets look at the original equation again
\(x+2y+3z=4\\
5x+4y+3z=8\)
We'll need to remove x to form some form of y+z, let's see if this is possible
take the first equation, multiply by 5
\(5x+10y+15z=20\\
minus\\
5x+4y+3z=8\\
=\\
6y+12z=12\\
=\\
y+2z=2\)
perfect! (Sometimes things don't work out this perfectly and we'll have to eliminate 1 by 1)
we have
\(2x+y=2\\
and\\
y+2z=2\)
add these together
\(2x+2y+2z=4\\
x+y+z=2\)
Both ways of solving are okay, the elimination may take longer on these perfectly lined up problems. But the short-cut cannot always be used. More method = more chance to get it right on gmat. Learn both if possible.
Option B