Question

10 which is the most efficient method for solving these systems of linear equations?

	substitution	graphing	elimination
	$x = 4y + 1$ <br> $2x - 3y = 12$	○	○	○
	$2x - 3y = 6$ <br> $x + 4y = -8$	○	○	○
	$5x = 3y - 7$ <br> $2y + 3 = 5x$	○	○	○

Explanation:

Step1: Analyze first system

First system: $y=\frac{1}{3}x+5$, $y=-2x-7$.
Both equations solve for $y$, so substitute one $y$ into the other directly. Substitution is efficient here.

Step2: Analyze second system

Second system: $x=4y+1$, $2x-3y=12$.
One equation solves for $x$, substitute $x$ into the second equation. Substitution is efficient here.

Step3: Analyze third system

Third system: $2x-3y=0$, $x+4y=-8$.
Equations are in standard form; coefficients of $x$ can be aligned for elimination, but substitution (solve for $x$ from second eq) also works. However, elimination is straightforward, but substitution is still simple. Wait, no: since one variable can be isolated easily, but elimination is also option. But actually, substitution is efficient here too? No, wait: for $2x-3y=0$ and $x+4y=-8$, solve for $x=-4y-8$, substitute into first equation: $2(-4y-8)-3y=0$. That's substitution. But elimination: multiply second equation by 2: $2x+8y=-16$, subtract first equation: $(2x+8y)-(2x-3y)=-16-0 \implies 11y=-16$. Either works, but substitution is direct as one variable has coefficient 1.

Wait no, let's re-express:
Wait the first two systems have one variable already isolated, so substitution is most efficient.
Third system: $2x-3y=0$, $x+4y=-8$: $x$ has coefficient 1, so solve for $x$ and substitute, substitution is efficient.
Fourth system: $5x=3y-7$, $2y+3=5x$. Both equal to $5x$, so substitute $5x$ from first into second: $2y+3=3y-7$. Substitution is efficient here as well? Wait no, elimination: subtract first from second: $(2y+3)-5x=(5x)-(3y-7)$? No, better to set equal: $3y-7=2y+3$, which is substitution (substitute $5x$).

Wait no, let's correct:

1. System 1: $y=\frac{1}{3}x+5$; $y=-2x-7$: Substitution (substitute one y into the other, no rearrangement needed)
2. System 2: $x=4y+1$; $2x-3y=12$: Substitution (substitute x into second equation directly)
3. System 3: $2x-3y=0$; $x+4y=-8$: Substitution (solve for $x=-4y-8$ easily, substitute into first equation)
4. System 4: $5x=3y-7$; $2y+3=5x$: Substitution (substitute $5x$ from first into second: $2y+3=3y-7$)

Wait, but elimination works for some, but substitution is more efficient here because at least one variable is isolated or can be isolated in 1 step, or expressions are equal to the same variable/term.

Answer:

1. System $\boldsymbol{y=\frac{1}{3}x+5; y=-2x-7}$: Substitution
2. System $\boldsymbol{x=4y+1; 2x-3y=12}$: Substitution
3. System $\boldsymbol{2x-3y=0; x+4y=-8}$: Substitution
4. System $\boldsymbol{5x=3y-7; 2y+3=5x}$: Substitution