a student is solving the system of equations below using the elimination method. system: $ 2x + 5y = 9 $ $ -2x + 3y = 15 $ the students work is shown. step 1: add the equations to eliminate $ x $ $ 4y = 24 $ step 2: solve for $ y $ $ y = 6 $ step 3: substitute $ y = 6 $ into the first equation $ 2x + 5(6) = 9 $ step 4: simplify $ 2x + 30 = 9 $ $ 2x = -21 $ $ x = -10.5 $ step 5: the student concludes the solution is $ (-10.5, 6) $ what error did the student make? a the student incorrectly added the two equations in step 1 b the student substituted the value for the wrong variable in step 3 c the student made an arithmetic error when solving for $ y $ d the student should have multiplied instead of adding in step 1

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# Brief Explanations:
First, let's analyze the elimination method steps. In Step 3, the student substitutes $ y = 8 $ into the "first equation", but when using elimination, after finding $ y $, we should substitute into one of the original equations (either of the two in the system) to find $ x $. Let's assume the original system is $ 2x + 5y = 3 $ and $ 2x - y = - 29 $ (since adding them would give $ 4y= - 26 $? Wait, no, maybe the original equations are $ 2x + 5y = 3 $ and $ - 2x + y = 29 $? Wait, the student's Step 1: added the equations to eliminate $ x $, getting $ 4y = 34 $? Wait, no, the student's Step 2: solved for $ y $, got $ y = 8 $. Wait, maybe the original equations are $ 2x + 5y = 3 $ and $ 2x - y = - 29 $. Adding them: $ (2x + 5y)+(2x - y)=3+( - 29)$ → $ 4x + 4y = - 26 $? No, that's not eliminating $ x $. Wait, maybe the original equations are $ 2x + 5y = 3 $ and $ - 2x + y = 29 $. Then adding: $ (2x - 2x)+(5y + y)=3 + 29 $ → $ 6y = 32 $? No, the student has $ 4y = 34 $, then $ y = 8 $ (but $ 34\div4 = 8.5 $, not 8). Wait, maybe the error is in Step 3: the student substituted $ y = 8 $ into the wrong equation. Let's check the options:

Option B says the student substituted the value for the wrong variable in Step 3. Wait, no, Step 3: substitute $ y = 8 $ into the first equation. Wait, maybe the first equation is $ 2x + 5y = 3 $, and when substituting $ y = 8 $, we get $ 2x + 40 = 3 $ → $ 2x = - 37 $ → $ x = - 18.5 $, but the student's Step 4: $ 12 + x = 1 $ → $ x = - 11 $. Wait, maybe the original first equation was $ 12 + y = 1 $? No, that doesn't make sense. Wait, the key is: in Step 3, the student should substitute $ y = 8 $ into one of the original equations (to find $ x $), but instead, maybe they used the wrong equation (like a non - original one) or substituted for the wrong variable. Wait, the options:

A: Incorrectly added in Step 1. If we check the addition, if the equations are, say, $ 2x + 5y = 3 $ and $ - 2x + y = 29 $, adding gives $ 6y = 32 $, not $ 4y = 34 $, so maybe A? No, wait the student's Step 2: $ y = 8 $ from $ 4y = 34 $ (but $ 34\div4 = 8.5 $, so C: arithmetic error in solving for $ y $? Wait, $ 4y = 34 $ → $ y = 34/4 = 8.5 $, but the student got $ y = 8 $, so C? But the options:

Wait, the problem is about the error the student made. Let's re - examine:

Step 1: Add equations to eliminate $ x $. Suppose the two equations are $ 2x + 5y = 3 $ and $ - 2x + y = 29 $. Adding them: $ (2x - 2x)+(5y + y)=3 + 29$ → $ 6y = 32 $, not $ 4y = 34 $. So maybe the equations are $ 2x + 3y = \dots $ (the image is a bit unclear). Wait, the student's Step 1: $ 4y = 34 $, Step 2: $ y = 8 $ (but $ 34\div4 = 8.5 $, so arithmetic error in solving for $ y $ (Option C)? But wait, the options:

Wait, the correct answer is B? No, let's think again. Wait, in Step 3, the student substitutes $ y = 8 $ into the "first equation", but maybe the first equation is not the one they should have used. Wait, no, the main error: when solving for $ y $ from $ 4y = 34 $, $ y = 34/4 = 8.5 $, but the student got $ y = 8 $, so that's an arithmetic error (Option C). But wait, the options:

Wait, the options are:

A: Incorrectly added in Step 1.

B: Substituted for wrong variable in Step 3.

C: Arithmetic error in solving for $ y $.

D: Should have multiplied instead of added in Step 1.

Let's assume the two equations are $ 2x + 5y = 3 $ and $ - 2x + y = 29 $. Adding them: $ (2x - 2x)+(5y + y)=3 + 29$ → $ 6y = 32 $, so if the student got $ 4y = 34 $, that's a wrong addition (A). But if the equations are $ 2x + 5y = 3 $ and $ 2x - y = - 29 $, adding them: $ 4x + 4y = - 26 $, which doesn't eliminate $ x $, so D: should have multiplied. But the student added to eliminate $ x $, so if the coefficients of $ x $ are equal and opposite, adding eliminates $ x $. Suppose the equations are $ 2x + 5y = 3 $ and $ - 2x + y = 29 $. Adding: $ 6y = 32 $ → $ y = 32/6 = 16/3\approx5.33 $. The student has $ 4y = 34 $ → $ y = 8.5 $. So maybe the equations are $ 2x + 5y = 3 $ and $ - 2x - y = - 29 $. Adding: $ 4y = 32 $ → $ y = 8 $. Ah! So if the second equation is $ - 2x - y = - 29 $, then adding $ 2x + 5y = 3 $ and $ - 2x - y = - 29 $ gives $ 4y = - 26 $? No, $ 3+( - 29)= - 26 $, so $ 4y = - 26 $ → $ y = - 6.5 $. Wait, I'm getting confused. Let's look at the options again.

Option B: "The student substituted the value for the wrong variable in Step 3". Step 3: substitute $ y = 8 $ into the first equation. If the first equation is, say, $ 12 + y = 1 $ (which doesn't make sense with the system), but maybe the student was supposed to substitute into an equation with $ x $ and $ y $, but instead substituted into an equation with only $ y $ or the wrong variable. Wait, maybe the original system is $ 2x + 5y = 3 $ and $ x - y = - 12 $ (just guessing). No, this is too unclear. Wait, the key is that in Step 3, when substituting $ y = 8 $ into the first equation, the student might have used the wrong equation (like a non - original one) or substituted for $ x $ instead of $ y $ or vice - versa. But the most probable error is:

Wait, the options:

A: Incorrect addition in Step 1.

B: Substituted for wrong variable in Step 3.

C: Arithmetic error in solving for $ y $.

D: Should have multiplied in Step 1.

Let's assume the two equations are $ 2x + 5y = 3 $ and $ - 2x + y = 29 $. Adding them: $ 6y = 32 $ → $ y = 32/6 = 16/3\approx5.33 $. The student has $ 4y = 34 $ → $ y = 8.5 $, so arithmetic error in Step 2 (solving for $ y $)? No, Step 2 is solving for $ y $ from $ 4y = 34 $, so $ y = 34\div4 = 8.5 $, but the student wrote $ y = 8 $, so C: arithmetic error when solving for $ y $. But maybe the correct answer is B: because in Step 3, the student should substitute $ y = 8 $ into an equation to find $ x $, but instead, maybe they substituted into an equation that doesn't have $ x $ or substituted for the wrong variable.

Wait, the problem is a bit unclear due to the image, but based on common errors in elimination:

If in Step 3, the student substitutes $ y = 8 $ into an equation that is not one of the original equations (like a miscalculated one) or substitutes for the wrong variable (e.g., substitutes $ y $ into an equation where they should substitute $ x $), but more likely, the error is in Step 3: substituting for the wrong variable. Wait, no, let's think again.

Suppose the original system is:

Equation 1: $ 2x + 5y = 3 $

Equation 2: $ - 2x + y = 29 $

Step 1: Add them: $ 6y = 32 $ → $ y = 32/6 = 16/3$

But the student did Step 1: $ 4y = 34 $ (wrong addition, so A), Step 2: $ y = 8 $ (arithmetic error, C), Step 3: substitutes $ y = 8 $ into the first equation.

But the options:

The correct answer is B: The student substituted the value for the wrong variable in Step 3. Wait, maybe the first equation is $ 12 + x = 1 $ (which is not related to the system), so when substituting $ y = 8 $ into it, it's the wrong variable (since the equation has $ x $, not $ y $). So the student should have substituted $ y = 8 $ into an equation with $ x $ and $ y $ (like Equation 1: $ 2x + 5y = 3 $) to find $ x $, but instead substituted into an equation with only $ x $ (like $ 12 + x = 1 $), which is the wrong variable (substituted $ y $ into an equation for $ x $). So the error is in Step 3: substituting for the wrong variable.

# Answer:
B. The student substituted the value for the wrong variable in Step 3

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