This question tests solving systems of two linear equations using substitution (replace variable), elimination (add/subtract to cancel variable), or graphing (estimate intersection). Elimination: align equations (multiply if needed), add/subtract to cancel variable ($2x+3y=13$ minus $2x-y=5$ gives $4y=8$), solve ($y=2$), substitute back ($x=3.5$). For this specific system, subtract the second equation from the first: ($2x+3y$) - ($2x - y$) = $13-5$, which is $4y=8$, so $y=2$, then plug into $2x - y=5$: $2x - 2=5$, $2x=7$, $x=3.5$. The correct solution is $ (3.5,2) $, and verify: $2(3.5)+3(2)=7+6=13$, and $2(3.5)-2=7-2=5$. A common error is wrong elimination operation, like adding instead of subtracting, which would give $4x+2y=18$ instead of canceling x. Choosing method: elimination is better here since x coefficients align (both 2), allowing easy subtraction to cancel x. Process: (1) choose method, (2) apply (substitute or eliminate), (3) solve one-variable equation, (4) back-substitute for second variable, (5) verify in both originals (both true confirms solution).

This question tests solving systems of two linear equations using substitution (replace variable), elimination (add/subtract to cancel variable), or graphing (estimate intersection), applied to a word problem about costs. Substitution: solve one equation for variable ($y=2x+1$), substitute into other ($3x+y=11$ becomes $3x+(2x+1)=11$), solve resulting one-variable equation ($5x=10 \rightarrow x=2$), back-substitute for other variable ($y=2(2)+1=5$). Elimination: align equations (multiply if needed), add/subtract to cancel variable ($2x+3y=13$ minus $2x-y=5$ gives $4y=8$), solve ($y=2$), substitute back ($x=3.5$). Both yield same solution ($x,y$) pair. For this system, multiply second by 2: $2x+4y=14$, subtract first: ($2x+4y$)-($2x+3y$)=14-11, $y=3$, then $x+2(3)=7$, $x=1$, giving $(1,3)$; verify: $2(1)+3(3)=2+9=11$, $1+6=7$. A common error is sign error in elimination or wrong multiplication. Process: (1) choose method, (2) apply (substitute or eliminate), (3) solve one-variable equation, (4) back-substitute for second variable, (5) verify in both originals (both true confirms solution).

This question tests solving systems of two linear equations using substitution (replace variable), elimination (add/subtract to cancel variable), or graphing (estimate intersection). Elimination: align equations (multiply if needed), add/subtract to cancel variable ($$2x+3y=13$$ minus $$2x-y=5$$ gives $$4y=8$$), solve ($$y=2$$), substitute back ($$x=3.5$$). For this specific system, add the equations: $$3x+2y + x-2y =16+0$$, so $$4x=16$$, $$x=4$$, then from $$x-2y=0$$: $$4-2y=0$$, $$y=2$$, giving $(4,2)$. The correct solution is $(4,2)$, verify: $$3(4)+2(2)=12+4=16$$, and $$4-2(2)=4-4=0$$. A common error is wrong elimination, like subtracting instead of adding, failing to cancel y. Choosing method: elimination is better when coefficients align oppositely, like +2y and -2y here for addition. Process: (1) choose method, (2) apply (substitute or eliminate), (3) solve one-variable equation, (4) back-substitute for second variable, (5) verify in both originals (both true confirms solution).

This question tests solving systems of two linear equations using substitution (replace variable), elimination (add/subtract to cancel variable), or graphing (estimate intersection). Substitution: solve one equation for variable ($y=2x+1$), substitute into other ($3x+y=11$ becomes $3x+(2x+1)=11$), solve resulting one-variable equation ($5x=10 \rightarrow x=2$), back-substitute for other variable ($y=2(2)+1=5$). For this specific system, substitute $y=x+4$ into $2x+y=10$: $2x+(x+4)=10$, $3x+4=10$, $3x=6$, $x=2$, $y=6$, giving $(2,6)$. The correct solution is $(2,6)$, verify: $6=2+4$, and $2(2)+6=4+6=10$. A common error is incomplete substitution, like forgetting +4, or coordinates reversed to $(6,2)$. Choosing method: substitution easiest when variable isolated ($y=x+4$ given, substitute immediately). Process: (1) choose method, (2) apply (substitute or eliminate), (3) solve one-variable equation, (4) back-substitute for second variable, (5) verify in both originals (both true confirms solution).

This question tests solving systems of two linear equations using substitution (replace variable), elimination (add/subtract to cancel variable), or graphing (estimate intersection), but here it's about classifying consistency. Substitution: solve one equation for variable ($y=2x+1$), substitute into other ($3x+y=11$ becomes $3x+(2x+1)=11$), solve resulting one-variable equation ($5x=10 \to x=2$), back-substitute for other variable ($y=2(2)+1=5$). Elimination: align equations (multiply if needed), add/subtract to cancel variable ($2x+3y=13$ minus $2x-y=5$ gives $4y=8$), solve ($y=2$), substitute back ($x=3.5$). Both yield same solution (x,y) pair. For this specific system, subtracting gives $0= -3$, which is false, indicating no solution (parallel lines). A common error is thinking it's infinite solutions if ignoring the constants. Process: (1) choose method, (2) apply (substitute or eliminate), (3) if contradiction like $0=3$, no solution; if identity like $0=0$, infinite; otherwise solve. Mistakes: distributing negatives wrong (sign errors), adding when should subtract (or vice versa).

This question tests solving systems of two linear equations using substitution (replace variable), elimination (add/subtract to cancel variable), or graphing (estimate intersection). Substitution: solve one equation for variable ($y=2x+1$), substitute into other ($3x+y=11$ becomes $3x+(2x+1)=11$), solve resulting one-variable equation ($5x=10 \to x=2$), back-substitute for other variable ($y=2(2)+1=5$). For this specific system, since y is already isolated in the first equation, substitute $y=2x+1$ into the second: $3x + (2x+1) =11$, which simplifies to $5x+1=11$, then $5x=10$, so $x=2$, and $y=5$. The correct solution is $(2,5)$, and you can verify by plugging back: for $y=2(2)+1=5$ checks, and $3(2)+5=6+5=11$ checks. A common error is incomplete substitution, like forgetting to add the +1, leading to wrong x values, or reversing coordinates to $(5,2)$. Choosing method: substitution is easiest here since y is isolated, making it quick to plug in. Process: (1) choose method, (2) apply (substitute or eliminate), (3) solve one-variable equation, (4) back-substitute for second variable, (5) verify in both originals (both true confirms solution).

This question tests solving systems of two linear equations using substitution (replace variable), elimination (add/subtract to cancel variable), or graphing (estimate intersection), but here it's about classifying consistency. Substitution: solve one equation for variable ($y=2x+1$), substitute into other ($3x+y=11$ becomes $3x+(2x+1)=11$), solve resulting one-variable equation ($5x=10 \to x=2$), back-substitute for other variable ($y=2(2)+1=5$). Elimination: align equations (multiply if needed), add/subtract to cancel variable ($2x+3y=13$ minus $2x-y=5$ gives $4y=8$), solve ($y=2$), substitute back ($x=3.5$). Both yield same solution (x,y) pair. For this system, substitute $y=3x-2$ into second (rearranged as $6x-2y-4=0$): $6x-2(3x-2)-4=6x-6x+4-4=0=0$, true for all x, so infinitely many solutions (same line). A common error is mistaking for no solution if sign error occurs. Process: (1) choose method, (2) apply (substitute or eliminate), (3) if identity like $0=0$, infinite solutions; if contradiction, none; otherwise one solution.

This question tests solving systems of two linear equations using substitution (replace variable), elimination (add/subtract to cancel variable), or graphing (estimate intersection). Substitution: solve one equation for variable ($y=2x+1$), substitute into other ($3x+y=11$ becomes $3x+(2x+1)=11$), solve resulting one-variable equation ($5x=10 \to x=2$), back-substitute for other variable ($y=2(2)+1=5$). Elimination: align equations (multiply if needed), add/subtract to cancel variable ($2x+3y=13$ minus $2x-y=5$ gives $4y=8$), solve ($y=2$), substitute back ($x=3.5$). Both yield same solution (x,y) pair. For this simple system, substitute x=6 into x+y=10: $6+y=10$, $y=4$, giving $(6,4)$. Verify: x=6, y=4 satisfies both obviously. A common error is coordinates reversed, like $(4,6)$. Choosing method: substitution easiest when variable isolated ($y=3x+1$ given, substitute immediately), elimination better when coefficients align ($2x+y=5$ and $2x-y=1$, subtract cancels $2x$). Process: (1) choose method, (2) apply (substitute or eliminate), (3) solve one-variable equation, (4) back-substitute for second variable, (5) verify in both originals (both true confirms solution). Mistakes: distributing negatives wrong (sign errors), adding when should subtract (or vice versa), finding x only without y, reversing coordinates.

This question tests solving systems of two linear equations using substitution (replace variable), elimination (add/subtract to cancel variable), or graphing (estimate intersection). Substitution: solve one equation for variable ($y=2x+1$), substitute into other ($3x+y=11$ becomes $3x+(2x+1)=11$), solve resulting one-variable equation ($5x=10 \rightarrow x=2$), back-substitute for other variable ($y=2(2)+1=5$). Elimination: align equations (multiply if needed), add/subtract to cancel variable ($2x+3y=13$ minus $2x-y=5$ gives $4y=8$), solve ($y=2$), substitute back ($x=3.5$). Both yield same solution ($x,y$) pair. For this system, use elimination as specified: subtract second from first ($2x+3y - (2x-y)=13-5$), so $4y=8$, $y=2$, then plug into $2x-y=5$ ($2x-2=5$, $2x=7$, $x=3.5$), giving $(3.5,2)$. Verify: for $x=3.5$, $y=2$ in first ($7+6=13$ true), second ($7-2=5$ true). A common error is wrong elimination operation, like adding instead of subtracting, leading to $2y=18$ or incorrect y. Choosing method: substitution easiest when variable isolated ($y=3x+1$ given, substitute immediately), elimination better when coefficients align ($2x+y=5$ and $2x-y=1$, subtract cancels $2x$). Process: (1) choose method, (2) apply (substitute or eliminate), (3) solve one-variable equation, (4) back-substitute for second variable, (5) verify in both originals (both true confirms solution). Mistakes: distributing negatives wrong (sign errors), adding when should subtract (or vice versa), finding x only without y, reversing coordinates.

This question tests solving systems of two linear equations using substitution (replace variable), elimination (add/subtract to cancel variable), or graphing (estimate intersection). Substitution: solve one equation for variable ($y=2x+1$), substitute into other ($3x+y=11$ becomes $3x+(2x+1)=11$), solve resulting one-variable equation ($5x=10 \rightarrow x=2$), back-substitute for other variable ($y=2(2)+1=5$). Elimination: align equations (multiply if needed), add/subtract to cancel variable ($2x+3y=13$ minus $2x-y=5$ gives $4y=8$), solve ($y=2$), substitute back ($x=3.5$). Both yield same solution ($x,y$) pair. For this system, use substitution since y is isolated: plug $y=2x+1$ into $3x+y=11$ to get $3x+2x+1=11$, so $5x=10$, $x=2$, then $y=5$, giving $(2,5)$. Verify: for $x=2$, $y=5$ in first ($5=4+1$ true), second ($6+5=11$ true). A common error is incomplete substitution, like forgetting the +1, leading to wrong x. Choosing method: substitution easiest when variable isolated ($y=2x+1$ given, substitute immediately), elimination better when coefficients align ($2x+y=5$ and $2x-y=1$, subtract cancels $2x$). Process: (1) choose method, (2) apply (substitute or eliminate), (3) solve one-variable equation, (4) back-substitute for second variable, (5) verify in both originals (both true confirms solution). Mistakes: distributing negatives wrong (sign errors), adding when should subtract (or vice versa), finding x only without y, reversing coordinates.