Question

1. solve each system of equations.

a. \\(\

$$\begin{cases} 2x + 3y = 4 \\\\ 2x = 7y + 24 \\end{cases}$$

\\)
b. \\(\

$$\begin{cases} 5x + 3y = 23 \\\\ 3y = 15x - 21 \\end{cases}$$

\\)

Explanation:

Part a

Step1: Substitute \(2x\) from the second equation into the first equation.

The second equation is \(2x = 7y + 24\). Substitute this into the first equation \(2x + 3y = 4\), we get \((7y + 24) + 3y = 4\).

Step2: Simplify and solve for \(y\).

Combine like terms: \(7y + 3y + 24 = 4\) → \(10y + 24 = 4\). Subtract 24 from both sides: \(10y = 4 - 24 = -20\). Divide both sides by 10: \(y = \frac{-20}{10} = -2\).

Step3: Substitute \(y = -2\) back into the second equation to find \(x\).

Substitute \(y = -2\) into \(2x = 7y + 24\): \(2x = 7(-2) + 24 = -14 + 24 = 10\). Divide both sides by 2: \(x = \frac{10}{2} = 5\).

Step1: Substitute \(3y\) from the second equation into the first equation.

The second equation is \(3y = 15x - 21\). Substitute this into the first equation \(5x + 3y = 23\), we get \(5x + (15x - 21) = 23\).

Step2: Simplify and solve for \(x\).

Combine like terms: \(5x + 15x - 21 = 23\) → \(20x - 21 = 23\). Add 21 to both sides: \(20x = 23 + 21 = 44\). Divide both sides by 20: \(x = \frac{44}{20} = \frac{11}{5} = 2.2\).

Step3: Substitute \(x = \frac{11}{5}\) back into the second equation to find \(y\).

Substitute \(x = \frac{11}{5}\) into \(3y = 15x - 21\): \(3y = 15(\frac{11}{5}) - 21 = 33 - 21 = 12\). Divide both sides by 3: \(y = \frac{12}{3} = 4\).

Answer:

The solution for part a is \(x = 5\), \(y = -2\)

Part b