Systems with More Variables: Practice

Subtract the first equation from the second to isolate $x$.

$(2x + y + z) - (x + y + z) = 7 - 6 \;\Rightarrow\; x = 1$

Subtracting the first from the third gives $y = 2$, and the first equation then gives $z = 3$, so the solution is $(1, 2, 3)$.

The second equation matches the first except for an extra $x$, so subtract them.

$(2x + y + z) - (x + y + z) = 9 - 6 \;\Rightarrow\; x = 3$

Subtract the first equation from the second to get $2x = 4$, so $x = 2$; subtract the first from the third to get $y = 2$. Then use the first equation.

$z = 5 - 2 - 2 = 1$

Subtract the first equation from the third; the $x$ and $z$ terms cancel.

$(x + 4y + z) - (x + y + z) = 9 - 6 \;\Rightarrow\; 3y = 3 \;\Rightarrow\; y = 1$

Subtract the second equation from the first to cancel $x$ and $z$.

$(x + y + z) - (x - y + z) = 6 - 0 \;\Rightarrow\; 2y = 6 \;\Rightarrow\; y = 3$

Subtracting the third from the first gives $z = 1$, and then $x = 2$, so the solution is $(2, 3, 1)$.

Add the first two equations; the $y$ and $z$ terms cancel.

$(x + y + z) + (x - y - z) = 7 + 1 \;\Rightarrow\; 2x = 8 \;\Rightarrow\; x = 4$

Subtract the second equation from the first; the $x$ and $y$ terms cancel.

$(x + y + z) - (x + y - z) = 8 - (-2) \;\Rightarrow\; 2z = 10 \;\Rightarrow\; z = 5$

Subtract the second equation from the first to get $2y = 4$, so $y = 2$; subtract the third from the first to get $2z = 2$, so $z = 1$. Then the first equation gives $x$.

$x = 6 - 2 - 1 = 3$

Subtract the first equation from the second to get $x = 2$, and from the third to get $z = 3$. Then the first equation gives $y$.

$y = 7 - 2 - 3 = 2$

Add the first two equations; the $y$ and $z$ terms cancel.

$(x + y + z) + (x - y - z) = 8 + 2 \;\Rightarrow\; 2x = 10 \;\Rightarrow\; x = 5$

Subtract the third equation from the first to get $2y = 2$, so $y = 1$. Adding the first and second gives $3x + 2y = 8$.

$3x + 2(1) = 8 \;\Rightarrow\; 3x = 6 \;\Rightarrow\; x = 2$

Subtract the third equation from the first to get $2y = 2$, so $y = 1$. Adding the first and second gives $3x + 2y = 11$.

$3x + 2(1) = 11 \;\Rightarrow\; 3x = 9 \;\Rightarrow\; x = 3$