How to Solve the System of Linear Equations 9x+11y=-9 and 3x+4y=-3

Question

Solve the system of equations: $\begin{cases}9x + 11y = -9 \\ 3x + 4y = -3\end{cases}$

Accepted Answer

$x=1, y=-\frac{3}{2}$

Answer

Step 1: Use the elimination method. Multiply the second equation $3x + 4y = -3$ by 3 to make the coefficient of $x$ match the first equation: $3\times(3x + 4y)=3\times(-3)$, which simplifies to $9x + 12y = -9$. Step 2: Subtract the first original equation $9x + 11y = -9$ from the new equation obtained in Step 1: $(9x + 12y)-(9x + 11y)=-9-(-9)$. This simplifies to $y=0$. Step 3: Substitute $y=0$ into the second original equation $3x + 4\times0 = -3$, which simplifies to $3x=-3$. Solve for $x$: $x=-1$. Step 4: Verify the solution by substituting $x=-1$ and $y=0$ into both original equations: For $9x + 11y$, we get $9\times(-1)+11\times0=-9$, which matches the right-hand side. For $3x + 4y$, we get $3\times(-1)+4\times0=-3$, which also matches the right-hand side. The solution is valid.