Simultaneous Equations: Elimination Methods

Feedback: Not quite! To solve this, start by adding the two equations together: $(x + y) + (x - y) = 8 + 6$. This eliminates $y$ and simplifies to $2x = 14$, so $x = 7$. Next, substitute $x = 7$ into either equation. Using $x + y = 8$, we get $7 + y = 8$, which simplifies to $y = 1$. Therefore, the solution is $x = 7$ and $y = 1$.

Feedback: Not quite! Add the two equations together: $(2x + y) + (x - y) = 10 + 2$. This eliminates $y$ and simplifies to $3x = 12$, so $x = 4$. Next, substitute $x = 4$ into either equation. Using $2x + y = 10$, we get $2(4) + y = 10$, which simplifies to $8 + y = 10$, so $y = 2$. Therefore, the solution is $x = 4$ and $y = 2$.

Feedback: Not quite! Add the two equations together: $(x + 2y) + (3x - 2y) = 10 + 6$. This eliminates $y$ and simplifies to $4x = 16$, so $x = 4$. Next, substitute $x = 4$ into either equation. Using $x + 2y = 10$, we get $4 + 2y = 10$, which simplifies to $2y = 6$, so $y = 3$. Therefore, the solution is $x = 4$ and $y = 3$.

Feedback: Not quite! Subtract the second equation from the first: $(3x + y) - (x + y) = 16 - 4$. This simplifies to $2x = 12$, so $x = 6$. Next, substitute $x = 6$ in $x + y = 4$, we get $6 + y = 4$, which simplifies to $y = -2$. Therefore, the solution is $x = 6$ and $y = -2$.

Feedback: Not quite! Subtract the first equation from the second: $(2x + 5y) - (2x + y) = 36 - 4$. This eliminates $2x$ and simplifies to $4y = 32$, so $y = 8$. Next, substitute $y = 8$ into $2x + y = 4$, we get $2x + 8 = 4$, which simplifies to $2x = -4$, so $x = -2$. Therefore, the solution is $x = -2$ and $y = 8$.

Feedback: Not quite! First multiply the second equation by 2 to align the coefficients of $y$ for elimination: $14x - 4y = 10$. Then, add the two equations together: $(6x + 4y) + (14x - 4y) = 10 + 10$, which simplifies to $20x = 20$, so $x = 1$. Substitute $x = 1$ into the first equation, $6x + 4y = 10$, to get $6 + 4y = 10$, which simplifies to $4y = 4$, so $y = 1$. Therefore, the solution is $x = 1$ and $y = 1$.