Question

find the solution of the system of equations.
$8x + 6y = 28$
$9x + 3y = 39$

Explanation:

Step1: Simplify the second equation

Multiply the second equation \(9x + 3y = 39\) by 2 to make the coefficients of \(y\) in both equations have a common multiple. We get \(18x + 6y = 78\).

Step2: Subtract the first equation from the new second equation

Subtract \(8x + 6y = 28\) from \(18x + 6y = 78\). So, \((18x + 6y)-(8x + 6y)=78 - 28\), which simplifies to \(10x=50\).

Step3: Solve for \(x\)

Divide both sides of \(10x = 50\) by 10. We get \(x=\frac{50}{10}=5\).

Step4: Substitute \(x = 5\) into the first equation

Substitute \(x = 5\) into \(8x+6y = 28\). So, \(8\times5+6y = 28\), which is \(40+6y = 28\).

Step5: Solve for \(y\)

Subtract 40 from both sides: \(6y=28 - 40=-12\). Then divide by 6: \(y=\frac{-12}{6}=-2\).

Answer:

The solution of the system is \(x = 5\) and \(y=-2\)