Question

$$2x-y+z=5; 4x+2y+3z=8; 3x-4y-z=3$$

Answer

x=1/3;y=-19/15;z=46/15

Solution


Solve for \(y\) in \(2x-y+z=5\).
\[y=-5+2x+z\]
Substitute \(y=-5+2x+z\) into \(4x+2y+3z=8\).
\[8x-10+5z=8\]
Substitute \(y=-5+2x+z\) into \(3x-4y-z=3\).
\[-5x+20-5z=3\]
Solve for \(x\) in \(8x-10+5z=8\).
\[x=\frac{18-5z}{8}\]
Substitute \(x=\frac{18-5z}{8}\) into \(y=-5+2x+z\).
\[y=-5+\frac{18-5z}{4}+z\]
Substitute \(x=\frac{18-5z}{8}\) into \(-5x+20-5z=3\).
\[-\frac{5(18-5z)}{8}+20-5z=3\]
Solve for \(z\) in \(-\frac{5(18-5z)}{8}+20-5z=3\).
\[z=\frac{46}{15}\]
Substitute \(z=\frac{46}{15}\) into \(y=-5+\frac{18-5z}{4}+z\).
\[y=-\frac{19}{15}\]
Substitute \(z=\frac{46}{15}\) into \(x=\frac{18-5z}{8}\).
\[x=\frac{1}{3}\]
Therefore,
\[\begin{aligned}&x=\frac{1}{3}\\&y=-\frac{19}{15}\\&z=\frac{46}{15}\end{aligned}\]