Question
Solve for x and y:
$\text{2x}+\text{5y}=\frac{8}{3},$
$3\text{x}-\text{2y}=\frac{5}{6}$
$\text{2x}+\text{5y}=\frac{8}{3},$
$3\text{x}-\text{2y}=\frac{5}{6}$
✓
Answer
The given equations are: $\text{2x}+\text{5y}=\frac{8}{3}\ \dots(1)$ $\text{3x}-\text{2y}=\frac{5}{6}\ \dots(2)$ Multiply (1) by 2 and (2) by 5 $\text{4x}+\text{10y}=\frac{16}{3}\ \dots(3)$ $\text{15x}-\text{10y}=\frac{25}{6}\ \dots(4)$Adding (3) from (4), we get
$19\text{x}=\frac{57}6{}$ $\Rightarrow\text{x}=\frac{57}{6\times19}=\frac{1}{2}$Substitution $\text{x}=\frac{1}{2}$ in (3), we get
$2\times\frac{1}{2}+\text{10y}=\frac{16}{3}$ $\text{10y}=\frac{16}{3}-2$ $\Rightarrow\text{10y}=\frac{10}{3}$ $\text{y}=\frac{10}{3\times10}=\frac{1}{3}$$\therefore$ Solution is $\text{x}=\frac{1}{2}$ and $\text{y}=\frac{1}{3}$
$19\text{x}=\frac{57}6{}$ $\Rightarrow\text{x}=\frac{57}{6\times19}=\frac{1}{2}$Substitution $\text{x}=\frac{1}{2}$ in (3), we get
$2\times\frac{1}{2}+\text{10y}=\frac{16}{3}$ $\text{10y}=\frac{16}{3}-2$ $\Rightarrow\text{10y}=\frac{10}{3}$ $\text{y}=\frac{10}{3\times10}=\frac{1}{3}$$\therefore$ Solution is $\text{x}=\frac{1}{2}$ and $\text{y}=\frac{1}{3}$
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