Question

2x + 3y =4
2x - 3y =8
find x and y by substitution​

Answer 1

Step-by-step explanation:

hope this will help you!!

Answer 2

Step-by-step explanation:

$\bf \large \underline{\red{Question : - }}$

• 2x + 3y =4 and x - 3y =8 Find x and y by substitution

$\bf \large \underline{\red{given : - }}$

• $\sf \: 2x \: + 3y =4 \: and \: x - 3y =8 \:$

$\bf \large \underline{\red{To \: Find: - }}$

• $\text{ \sf \: find x and y by substitution}$

$\bf \large \underline{\red{solution: - }}$

$\sf \: 2x \: + 3y =4 \: eq \: (1) \\ \sf \:\: x - 3y =8 \: eq \: (2)$

Consider the first equation as Eq. (1) and} second equation as Eq. (2)

$\sf \: Considering \: Eq \: 1, \: we \: get: \\ \\ \sf \: 2x \: + 3y =4 \\ \\ \sf \: 2x = 4 - 3y \\ \\ \sf \: x = \frac{4 - 3y}{2} \: eq.(3)$

$\text{ \sf \red{Substituting this value of x in Eqn. 2, we get}}$

$\sf \to \: \: \: x - 3y =8 \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf \to \: \: \: \: \frac{4 - 3y}{2} - 3y = 8 \\ \\ \sf \red{ Taking \: LCM \: and \: solving,} \\ \\ \sf \to \: \: \: \frac{4 - 3y - 6y}{2} = 8 \\ \\ \sf \red{Transposing \: 2 \: to \: the \: RHS,} \\ \\ \sf \to \: \: 4 - 3y - 6y = 16 \\ \\ \sf \to \: \: \: - 9y = 16 - 4 \\ \\ \sf \to \: \: \: - 9y = 12 \\ \\ \sf \to \: \: \: y \: = - \frac{12}{9} \\ \\ \sf \to \: \: \: y \: = - \frac{4}{3}$

$\text{ \sf \red{Substituting value of 'y' in Eqn 3, we get:}}$

$\sf \to \: \: \: x \: = \frac{4 - 3 \times \frac{ - 4}{3} }{2} \\ \\ \sf \to \: \: \: x \: = \frac{4 + \frac{12}{3} }{2} \\ \\ \sf \to \: \: \: x \: = \frac{4 + 4}{2} \\ \\ \sf \to \: \: \: x = \frac{8}{2} \\ \\ \sf \to \red{x = 4}$

$\sf \pink{the \: value \: of \: x \: = 4 \: and\: y \: = \frac{ - 4}{3} }$

$\bf \large \dag \: \underline{\red{verification : - }} \:$

$\sf \to \: \: \: 2x + 3y = 4 \\ \\ \sf \to \: \: \: 2 \times 4 + 3 \times - \frac{ 4}{3} = 4 \\ \\ \sf \to \: \: \: 8 - 4 = 4 \\ \\ \sf \to \: \: \: \: \: 0 = 0$