Question

1. $12x + 9y = 0$\\$(3x + 5y = 11)$ $(-4)$\\12) $8x + 8y = 8$\\$-7x - 16y = 2$

Explanation:

Problem 10: Solve the system \(

$$\begin{cases}12x + 9y = 0\\3x + 5y = 11\end{cases}$$

\) (and the point \((-4)\) seems like a typo, maybe a miscalculation step or extra, we focus on the system)

Step1: Simplify the first equation

Divide the first equation \(12x + 9y = 0\) by 3, we get \(4x + 3y = 0\), then express \(x\) in terms of \(y\): \(4x=-3y\), so \(x = -\frac{3}{4}y\)

Step2: Substitute \(x = -\frac{3}{4}y\) into the second equation

Substitute into \(3x + 5y = 11\), we have \(3\times(-\frac{3}{4}y)+5y = 11\)
Calculate: \(-\frac{9}{4}y + 5y = 11\), combine like terms: \(\frac{-9 + 20}{4}y = 11\), that is \(\frac{11}{4}y = 11\)

Step3: Solve for \(y\)

Multiply both sides by \(\frac{4}{11}\), we get \(y = 11\times\frac{4}{11}=4\)

Step4: Solve for \(x\)

Substitute \(y = 4\) into \(x = -\frac{3}{4}y\), we get \(x = -\frac{3}{4}\times4=-3\)

Step1: Simplify the first equation

Divide the first equation \(8x + 8y = 8\) by 8, we get \(x + y = 1\), then express \(x\) in terms of \(y\): \(x = 1 - y\)

Step2: Substitute \(x = 1 - y\) into the second equation

Substitute into \(-7x - 16y = 2\), we have \(-7(1 - y)-16y = 2\)

Step3: Expand and solve for \(y\)

Expand: \(-7 + 7y - 16y = 2\), combine like terms: \(-7 - 9y = 2\)
Add 7 to both sides: \(-9y = 2 + 7 = 9\)
Divide by \(-9\): \(y = -1\)

Step4: Solve for \(x\)

Substitute \(y = -1\) into \(x = 1 - y\), we get \(x = 1 - (-1)=2\)

Answer:

The solution is \(x = -3\), \(y = 4\) (or the ordered pair \((-3, 4)\))

Problem 12: Solve the system \(

$$\begin{cases}8x + 8y = 8\\-7x - 16y = 2\end{cases}$$

\)