This system tests solving two linear equations using substitution (replace variable), elimination (add/subtract to cancel variable), or graphing (estimate intersection). The second equation directly gives x = 6, making this a simple substitution problem: substitute x = 6 into the first equation to get $6 + y = 10$, which gives $y = 4$. The solution is $(6, 4)$, verified by checking: $6 + 4 = 10$ ✓ and $x = 6$ ✓. Common mistakes include reversing the coordinates to get (4, 6) or misreading which variable is given. When one equation provides a variable's value directly, always use substitution for efficiency.

This system tests solving two linear equations using substitution (replace variable), elimination (add/subtract to cancel variable), or graphing (estimate intersection). The second equation can be rewritten as $ y = 2x + 3 $ by dividing both sides by 2, which is identical to the first equation. Since both equations represent the same line, every point on the line $ y = 2x + 3 $ is a solution, giving infinitely many solutions. This can be verified algebraically: the second equation $ 2y = 4x + 6 $ is exactly twice the first equation $ y = 2x + 3 $. Students might confuse this with no solution (parallel lines) or think there's only one solution. When equations are multiples of each other, the system has infinitely many solutions.

This system tests solving two linear equations using substitution (replace variable), elimination (add/subtract to cancel variable), or graphing (estimate intersection). Since the first equation already has $y$ isolated ($y = 2x + 1$), substitution is the most efficient method: substitute this expression into the second equation to get $3x + (2x + 1) = 11$, which simplifies to $5x + 1 = 11$, then $5x = 10$, so $x = 2$. Back-substituting $x = 2$ into $y = 2x + 1$ gives $y = 2(2) + 1 = 5$, so the solution is $(2, 5)$. To verify, check both equations: $y = 2(2) + 1 = 5$ ✓ and $3(2) + 5 = 11$ ✓. Common errors include reversing the coordinates to get $(5, 2)$ or making arithmetic mistakes when combining like terms.

This system tests solving two linear equations using substitution (replace variable), elimination (add/subtract to cancel variable), or graphing (estimate intersection). Elimination works well here since both equations have the same x-coefficient (2x): subtract the second equation from the first to eliminate x, giving $(2x + 3y) - (2x - y) = 13 - 5$, which simplifies to $4y = 8$, so $y = 2$. Substitute $y = 2$ back into either original equation: using $2x - y = 5$, we get $2x - 2 = 5$, so $2x = 7$, and $x = 3.5$. The solution is $(3.5, 2)$, which can be verified: $2(3.5) + 3(2) = 7 + 6 = 13$ ✓ and $2(3.5) - 2 = 7 - 2 = 5$ ✓. Students might reverse coordinates or make sign errors when subtracting equations.

This system tests solving two linear equations using substitution (replace variable), elimination (add/subtract to cancel variable), or graphing (estimate intersection). Adding the equations eliminates y: (x + y) + (2x - y) = 7 + 5, which gives $3x = 12$, so $x = 4$. Substitute x = 4 into the first equation: $4 + y = 7$, so $y = 3$, giving the solution $(4, 3)$. Verification confirms both equations: $4 + 3 = 7$ ✓ and $2(4) - 3 = 8 - 3 = 5$ ✓. Common mistakes include subtracting instead of adding (which would give $x = \frac{2}{3}$) or reversing the coordinates. When y-coefficients are opposites, addition eliminates y efficiently.

This question tests solving systems of two linear equations using substitution (replace variable), elimination (add/subtract to cancel variable), or graphing (estimate intersection), but specifies substitution. 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, using substitution, replace y in the second equation with 2x + 1 to get $3x + 2x + 1 = 11$, simplify to $5x = 10$, so $x = 2$, then $y = 5$. The correct solution is (2,5), and verification shows it satisfies both equations. A common error is forgetting to add the +1 during substitution or reversing the coordinates.

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). 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 system tests solving two linear equations using substitution (replace variable), elimination (add/subtract to cancel variable), or graphing (estimate intersection). Both equations have the form $2x + y = \text{constant}$, but with different constants ($5$ and $8$), meaning they represent parallel lines with the same slope but different y-intercepts. Since parallel lines never intersect, this system has no solution. To verify algebraically: if we subtract the first equation from the second, we get $0 = 3$, which is a contradiction. Students might incorrectly think the system has infinitely many solutions (which occurs when equations are identical) or try to force a solution. Recognizing parallel lines (same coefficients, different constants) immediately identifies no 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).