Question

Find the value of a and b using substitution method:
$\left(\frac{1}{2}\right)a+\left(\frac{1}{2}\right)b=3$ and $a+2b=\frac{2}{3}$

• A. $\frac{34}{3}$ and $\frac{-16}{3}$
• B. $\frac{34}{3}$ and $\frac{16}{3}$
• C. $\frac{2}{3}$ and $\frac{-16}{3}$
• D. $\frac{-34}{3}$ and $\frac{16}{3}$

Answer 1

Answer: (A)

$\frac{34}{3}$ and $\frac{-16}{3}$

$\left(\frac{1}{2}\right)a+\left(\frac{1}{2}\right)b=3$ ------- (1)
$a+2b=\frac{2}{3}$ ------ (2)
From equation 1:
$\frac{a}{2}+\frac{b}{2}=3$
$a+b=6$
$b=6-a$
Substitute the value of b in equation 2:
$a+2b=\frac{2}{3}$
$a+2\left(6a\right)=\frac{2}{3}$
$a+122a=\frac{2}{3}$
$-a=\frac{2}{3}-12$
$-a=\frac{\left(236\right)}{3}$
$-a=\frac{-34}{3}$
$a=\frac{34}{3}$
Now, Substitute $a=\frac{34}{3}$ in equation 1:
$a+2b=\frac{2}{3}$
$\frac{34}{3}+2b=\frac{2}{3}$
$\frac{(34+6b)}{3}=\frac{2}{3}$
$34+6b=2$
$6b=234$
$6b=-32$
$b=\frac{-16}{3}$
Therefore the solution is: $a=\frac{34}{3}$ and $b=\frac{-16}{3}$