Question

x/6 + y/15 = 4 and x/3 - y/12 = 9/4 solve for x and y

Answer 1

CROSS - MULTIPLICATION METHOD:

The general form of a pair of linear equations

a1x + b1y + c1 = 0 , & a2x + b2y + c2 = 0.

When a1 / a2 ≠ b1 / b2, the pair of linear equations will have a unique solution.

To solve this pair of equations for x and y using cross-multiplication, we’ll arrange the variables x and y and their coefficients a1, a2, b1 and b2, and the constants c1 and c2 as shown below

⇒ x = b1 c2 - b2 c1 / a1 b2 - a2 b1

⇒ y = c1 a2 - c2 a1 / a1 b2 - a2 b1

The above equation is generally written as :

x/ b1 c2 - b2 c1 = y/ c1 a2 - c2 a1 = 1/a1 b2 - a2 b1

For more details look at the attachment above :)

Answer 2

Step-by-step explanation:

$\frac{x}{6} + \frac{y}{15} = 4 \\ => \frac{5x + 2y}{30} = 4 \\ => 5x + 2y = 4 \times 30 \\ => 5x = 120 - 2y \\ => x = \frac{120 - 2y}{5} ............. \: (i)$

$\frac{x}{3} - \frac{y}{12} = \frac{9}{4} \\ => \frac{4x - y}{12} = \frac{9}{4} \\ => \frac{4x - y}{3} = 9 \\ => 4x - y = 9 \times 3 \\ => 4x = 27 + y \\ => x = \frac{27 + y}{4} ............... \: (ii)$

On equating both the equations (i) and (ii), we get,

$\frac{27 + y}{4} = \frac{120 - 2y}{5} \\ => 5(27 + y) = 4(120 - 2y) \\ => 135 + 5y = 480 - 8y \\ = > 5y + 8y = 480 - 135 \\ = > 13y = 345 \\ = > y = \frac{345}{13}$

So, y = $\frac{345}{13}$

So, x = $\frac{27 + y}{4} \\ = \frac{27 + \frac{345}{13} }{4} \\ = \frac{299 + 345}{13 \times 4} \\ = \frac{644}{52} \\ = \frac{161}{13}$