Question
Solve the following systems of equations:
$\frac{4}{\text{x}}+5\text{y}=7,$
$\frac{3}{\text{x}}+4\text{y}=5.$
$\frac{4}{\text{x}}+5\text{y}=7,$
$\frac{3}{\text{x}}+4\text{y}=5.$
✓
Answer
The given equations are
$\frac{4}{\text{x}}+5\text{y}=7\ ......(\text{i})$
$\frac{3}{\text{x}}+4\text{y}=5\ ......(\text{ii})$
Let $\frac{1}{\text{x}}=\text{u}$ so,
$4\text{u}+5\text{y}=7\ ....(\text{iii})$
$3\text{u}+4\text{y}=5\ .....(\text{iv})$
Multiplying (iii) by 3 and (iv) by 4, we get
$\Rightarrow12\text{u}+15\text{y}=21\ ......(\text{v})$
$\Rightarrow12\text{u}+16\text{y}=20\ .......(\text{vi})$
Subtracting (v) from (vi), we get
$\Rightarrow1\text{y}=-1$
$\Rightarrow\text{y}=-1$
Putting y = -1 in (iv) we get
$\Rightarrow3\text{u}+4(-1)=5$
$\Rightarrow3\text{u}=9$
$\Rightarrow\text{u}=3$
$\therefore\frac{1}{\text{x}}=\text{u}$
$\Rightarrow\text{x}=\frac{1}{\text{u}}\Rightarrow\text{x}=\frac{1}{3}$
Thus, $\text{x}=\frac{1}{3}$ and y = -1.
$\frac{4}{\text{x}}+5\text{y}=7\ ......(\text{i})$
$\frac{3}{\text{x}}+4\text{y}=5\ ......(\text{ii})$
Let $\frac{1}{\text{x}}=\text{u}$ so,
$4\text{u}+5\text{y}=7\ ....(\text{iii})$
$3\text{u}+4\text{y}=5\ .....(\text{iv})$
Multiplying (iii) by 3 and (iv) by 4, we get
$\Rightarrow12\text{u}+15\text{y}=21\ ......(\text{v})$
$\Rightarrow12\text{u}+16\text{y}=20\ .......(\text{vi})$
Subtracting (v) from (vi), we get
$\Rightarrow1\text{y}=-1$
$\Rightarrow\text{y}=-1$
Putting y = -1 in (iv) we get
$\Rightarrow3\text{u}+4(-1)=5$
$\Rightarrow3\text{u}=9$
$\Rightarrow\text{u}=3$
$\therefore\frac{1}{\text{x}}=\text{u}$
$\Rightarrow\text{x}=\frac{1}{\text{u}}\Rightarrow\text{x}=\frac{1}{3}$
Thus, $\text{x}=\frac{1}{3}$ and y = -1.
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