Question
Solve the following systems of equations has unique solution and solve it:
$\frac{\text{x}}{3}+\frac{\text{y}}2{}=3,\ \text{x}-\text{2y}=2$
$\frac{\text{x}}{3}+\frac{\text{y}}2{}=3,\ \text{x}-\text{2y}=2$
✓
Answer
$\frac{\text{x}}{3}+\frac{\text{y}}{2}=3$
$\Rightarrow\ \frac{2\text{x}+3\text{y}}{6}=3$
$2x + 3y - 18 = 0 ...(1)$
$x - 2y - 2 = 0 ...(2)$
$a_1= 2, b_1 = 3, c_1 = -18$
$a_2 = 1, b_2 = -2, c_2 = -2$
Thus, $\frac{\text{a}_1}{\text{a}_2}\neq\frac{\text{b}_1}{\text{b}_2}\Rightarrow\ \frac{2}{1}\neq\frac{3}{-2}$
Hence, the given system of equations has unique solution.
Then given equations are
$2x + 3y = 18 ...(1)$
$x - 2y = 2 ...(2)$
Multiplying (1) by 2 and (2) by 3
$4x + 6y = 36 ...(3)$
$3x - 6y = 6 ...(4)$
Adding (3) and (4) we get
$7x = 42$
$\Rightarrow x = 6$
Putting x = 6 in (1), we get
$2 \times 6 + 3y = 18$
$\Rightarrow 3y = 18 - 12$
$\Rightarrow 3y = 6$
$\Rightarrow\text{y}=\frac{6}{3}=2$
$\therefore$ Solution is x = 6, y = 2.
$\Rightarrow\ \frac{2\text{x}+3\text{y}}{6}=3$
$2x + 3y - 18 = 0 ...(1)$
$x - 2y - 2 = 0 ...(2)$
$a_1= 2, b_1 = 3, c_1 = -18$
$a_2 = 1, b_2 = -2, c_2 = -2$
Thus, $\frac{\text{a}_1}{\text{a}_2}\neq\frac{\text{b}_1}{\text{b}_2}\Rightarrow\ \frac{2}{1}\neq\frac{3}{-2}$
Hence, the given system of equations has unique solution.
Then given equations are
$2x + 3y = 18 ...(1)$
$x - 2y = 2 ...(2)$
Multiplying (1) by 2 and (2) by 3
$4x + 6y = 36 ...(3)$
$3x - 6y = 6 ...(4)$
Adding (3) and (4) we get
$7x = 42$
$\Rightarrow x = 6$
Putting x = 6 in (1), we get
$2 \times 6 + 3y = 18$
$\Rightarrow 3y = 18 - 12$
$\Rightarrow 3y = 6$
$\Rightarrow\text{y}=\frac{6}{3}=2$
$\therefore$ Solution is x = 6, y = 2.
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