Question

Solve the following equations: \[ \dfrac{3}{2}x - \dfrac{5}{3}y = -2, \quad \dfrac{x}{3} + \dfrac{y}{2} = \dfrac{13}{6} \]

Hint:

To eliminate fractions, always multiply through by the LCM of denominators before applying elimination or substitution.

Answer 1

Step 1: Eliminate the fractions by taking the LCM of denominators. 
For the first equation: \[ \dfrac{3}{2}x - \dfrac{5}{3}y = -2 \] Multiply both sides by 6 (LCM of 2 and 3): \[ 9x - 10y = -12 \quad \text{(i)} \] For the second equation: \[ \dfrac{x}{3} + \dfrac{y}{2} = \dfrac{13}{6} \] Multiply both sides by 6: \[ 2x + 3y = 13 \quad \text{(ii)} \] Step 2: Solve the two linear equations. 
Equation (i): \(9x - 10y = -12\) Equation (ii): \(2x + 3y = 13\) 
Step 3: Multiply (ii) by 3 to align coefficients of \(y\). 
\[ 6x + 9y = 39 \quad \text{(iii)} \] Step 4: Eliminate \(y\). 
Multiply (i) by 3: \[ 27x - 30y = -36 \quad \text{(iv)} \] Multiply (ii) by 10: \[ 20x + 30y = 130 \quad \text{(v)} \] Add (iv) and (v): \[ 47x = 94 \Rightarrow x = 2 \] Step 5: Substitute \(x = 2\) in equation (ii). 
\[ 2(2) + 3y = 13 \Rightarrow 4 + 3y = 13 \Rightarrow 3y = 9 \Rightarrow y = 3 \] Step 6: Conclusion. 
Hence, \(x = 2\) and \(y = 3\).