Question

1. solve the simultaneous equations

$7a + 3b = 51$
$4a + b = 22$

Explanation:

Step1: Solve for \( b \) from the second equation

From \( 4a + b = 22 \), we can express \( b \) as \( b = 22 - 4a \).

Step2: Substitute \( b \) into the first equation

Substitute \( b = 22 - 4a \) into \( 7a + 3b = 51 \):
\[

$$\begin{align*} 7a + 3(22 - 4a) &= 51\\ 7a + 66 - 12a &= 51\\ -5a + 66 &= 51\\ -5a &= 51 - 66\\ -5a &= -15\\ a &= \frac{-15}{-5}\\ a &= 3 \end{align*}$$

\]

Step3: Substitute \( a = 3 \) back to find \( b \)

Substitute \( a = 3 \) into \( b = 22 - 4a \):
\( b = 22 - 4\times3 = 22 - 12 = 10 \)

Answer:

\( a = 3 \), \( b = 10 \)