Question

1. $2x + 5y = 11$; $-3x + 8y = -1$
2. $7x - 3y = 2$; $5x + 4y = -17$
3. $4x - 5y = -28$; $-9x - 2y = 10$
4. $2x + 3y = 10$; $3x - 10y = 15$
5. $-7x + 4y = -6$; $2x - 5y = 21$
6. $8x + 3y = -12$; $6x + 5y = -20$
7. $-4x - 9y = 1$; $-x + 2y = -4$
8. $5x - 12y = -16$; $-3x + 4y = 0$

Explanation:

Let's solve problem 5: \(

$$\begin{cases}2x + 5y = 11\\-3x + 8y = -1\end{cases}$$

\) using the elimination method.

Step 1: Eliminate \(x\)

Multiply the first equation by \(3\) and the second equation by \(2\) to make the coefficients of \(x\) opposite:

• First equation: \(3(2x + 5y) = 3\times11\) gives \(6x + 15y = 33\)
• Second equation: \(2(-3x + 8y) = 2\times(-1)\) gives \(-6x + 16y = -2\)

Step 2: Add the two new equations

\((6x + 15y) + (-6x + 16y) = 33 + (-2)\)
Simplify: \(31y = 31\)

Step 3: Solve for \(y\)

Divide both sides by \(31\): \(y = \frac{31}{31} = 1\)

Step 4: Substitute \(y = 1\) into the first original equation

\(2x + 5(1) = 11\)
Simplify: \(2x + 5 = 11\)
Subtract \(5\) from both sides: \(2x = 11 - 5 = 6\)
Divide by \(2\): \(x = \frac{6}{2} = 3\)

Answer:

The solution for problem 5 is \(x = 3\), \(y = 1\)