Solve a pair of substitution systems of equations

Key concept: Substitution is one of those methods that feels basic, but it stays useful because it keeps the algebra organized. Once one equation is already solved for a variable, the rest becomes straight replacement and cleanup.

Step by step

You actually have two separate systems here.

System 1

$y=6x+7$ $y=12x+5$

Since both equal $y$, set them equal to each other: $6x+7=12x+5.$ Solve: $7=6x+5$ $2=6x$ $x=\frac13.$ Now substitute back: $y=6\left(\frac13\right)+7=2+7=9.$

So the first system has solution $\boxed{\left(\frac13,9\right)}.$

System 2

$x=2y+1$ $y+3z+5=0$ $x-3z=0$

From $x-3z=0$, $x=3z.$ Set this equal to $2y+1$: $3z=2y+1.$ So $2y=3z-1,$ $y=\frac{3z-1}{2}.$ Substitute into $y+3z+5=0$: $\frac{3z-1}{2}+3z+5=0.$ Multiply by 2: $3z-1+6z+10=0$ $9z+9=0$ $z=-1.$ Then $x=3z=-3,$ and $y=2(-1)+?$ Better to use $x=2y+1$ with $x=-3$: $-3=2y+1$ $2y=-4$ $y=-2.$

So the second system has solution $\boxed{(x,y,z)=(-3,-2,-1)}.$

Pitfall alert

A common mistake is treating both groups of equations as one big system. They are separate problems, so solve each set on its own. Another easy error is mixing up the variable you are substituting into; keep track of whether you are replacing $x$, $y$, or $z$.

Try different conditions

If the first system had been $y=6x-7$ and $y=12x+5$, the same substitution step would work, but the final numbers would change. For the three-variable system, if one equation were replaced by a linearly dependent version, you might get infinitely many solutions instead of a single ordered triple.

Further reading

substitution method, system of equations, ordered triple