Questions: -3x-5y=13 -x+4y=10 and (-6,1)

Solution Steps

To solve the system of linear equations, we can use the method of substitution or elimination. Here, we will use the elimination method to find the values of \(x\) and \(y\). We will then check if the point \((-6, 1)\) satisfies both equations.

1. Multiply the second equation by 3 to align the coefficients of \(x\).
2. Add the two equations to eliminate \(x\) and solve for \(y\).
3. Substitute the value of \(y\) back into one of the original equations to solve for \(x\).
4. Verify if the point \((-6, 1)\) satisfies both equations.

We are given the following system of linear equations: \[ \begin{align_} -3x - 5y &= 13 \quad \text{(1)} \\ -x + 4y &= 10 \quad \text{(2)} \end{align_} \]

To eliminate \(x\), we can multiply equation (2) by 3: \[ \begin{align_} -3x + 12y &= 30 \quad \text{(3)} \end{align_} \] Now, we can add equations (1) and (3): \[ (-3x - 5y) + (-3x + 12y) = 13 + 30 \] This simplifies to: \[ -6x + 7y = 43 \]

From equation (3), we can express \(x\) in terms of \(y\): \[ -3x = 30 - 12y \implies x = -10 + 4y \] Substituting \(x\) back into equation (1): \[ -3(-10 + 4y) - 5y = 13 \] This simplifies to: \[ 30 - 12y - 5y = 13 \implies 30 - 17y = 13 \] Solving for \(y\): \[ -17y = 13 - 30 \implies -17y = -17 \implies y = 1 \] Now substituting \(y = 1\) back into the expression for \(x\): \[ x = -10 + 4(1) = -10 + 4 = -6 \]

We need to check if the point \((-6, 1)\) satisfies both original equations:

1. For equation (1): \[ -3(-6) - 5(1) = 18 - 5 = 13 \quad \text{(True)} \]
2. For equation (2): \[ -(-6) + 4(1) = 6 + 4 = 10 \quad \text{(True)} \]

Final Answer