Question

1. \\(-4x - 3y = -15\\)\\(4x - 7y = -15\\)\\(\\)\\(6) \\(3x + y = 20\\)\\(x + y = 12\\)

Explanation:

Problem 5: Solve the system of equations \(

$$\begin{cases}-4x - 3y=-15\\4x - 7y=-15\end{cases}$$

\)

Step1: Eliminate \(x\) by adding the two equations

Add the left - hand sides and the right - hand sides of the two equations respectively.
\((-4x - 3y)+(4x - 7y)=-15+(-15)\)
Simplify the left - hand side: \(-4x+4x-3y - 7y=-10y\)
Simplify the right - hand side: \(-30\)
So we get the equation \(-10y=-30\)

Step2: Solve for \(y\)

Divide both sides of the equation \(-10y = - 30\) by \(-10\):
\(y=\frac{-30}{-10}=3\)

Step3: Substitute \(y = 3\) into one of the original equations to solve for \(x\)

Substitute \(y = 3\) into the first equation \(-4x-3y=-15\):
\(-4x-3\times3=-15\)
\(-4x - 9=-15\)
Add 9 to both sides: \(-4x=-15 + 9=-6\)
Divide both sides by \(-4\): \(x=\frac{-6}{-4}=\frac{3}{2}\)

Problem 6: Solve the system of equations \(

$$\begin{cases}3x + y=20\\x + y=12\end{cases}$$

\)

Step1: Eliminate \(y\) by subtracting the two equations

Subtract the second equation from the first equation:
\((3x + y)-(x + y)=20 - 12\)
Simplify the left - hand side: \(3x + y-x - y = 2x\)
Simplify the right - hand side: \(8\)
So we get the equation \(2x=8\)

Step2: Solve for \(x\)

Divide both sides of the equation \(2x = 8\) by 2:
\(x=\frac{8}{2}=4\)

Step3: Substitute \(x = 4\) into one of the original equations to solve for \(y\)

Substitute \(x = 4\) into the second equation \(x + y=12\):
\(4 + y=12\)
Subtract 4 from both sides: \(y=12 - 4 = 8\)

Answer:

s:

• For problem 5: \(x=\frac{3}{2},y = 3\)
• For problem 6: \(x = 4,y = 8\)