Question

the soln of the pair of the linear equation 3x-4y+11=0 and 7x+11y+15=0​

Answer 1

$\red{ \underline \bold{question : 3x - 4y + 11 = 0}} \: \\ \\ \orange{ \underline \bold{solution : : : : : > }} \\ \\ \orange{ \underline \bold{subtract \: from \: both \: sides : : : : : > }} \\ \\ \tt : \implies \orange{ \underline \bold{(3x - 4y + 11) - 11 = 0 - 11}} \\ \\ \orange{ \underline \bold{remove \: (0)}} \\ \\ \tt : \implies \:\orange{ \underline \bold{3x - 4y = - 11}}\\ \\ \orange{ \underline \bold{add \: to \: both \: sides}} \\ \tt : \implies\orange{ \underline \bold{(3x - 4y) + 4y = - 11 + 4y}} \\ \\ \orange{ \underline \bold{simpify \: left \: side}} \\ \tt : \implies \: \orange{ \underline \bold{3x = - 11 + 4y}} \\ \\ \orange{ \underline \bold{divide \: both \: sides}} \\ \tt : \implies \: \orange{ \underline \bold{ \frac{3x}{3} = \frac{ - 11 + 4y}{3} }} \\ \\ \orange{ \underline \bold{simplify \: fraction}} \\ \tt : \implies \: \orange{ \underline \bold{ x = \frac{ - 11 + 4y}{3} }} \\ \\ \orange{ \underline \bold{brack \: up \: fraction}} \\ \\ \tt : \implies \: \orange{ \underline \bold {x = - \frac{ 11}{3} + 4 \times \frac{y}{3} }}answer......... \\ \\ \\ \\ \\ \red{ \underline \bold{ \: question : : : : > 7x + 11y + 15 = 0}} \\ \\ \orange{ \underline \bold{solution}} \\ \\ \orange{ \underline \bold{subtract \: 11y \: from \: both \: sides \: and \: remove(0)}} \\ \\ \tt : \implies \: \orange{ \underline \bold{7x + 15 = - 11y}} \\ \\ \orange{ \underline \bold{subtract \: 15 \: from \: both \: sides}} \\ \\ \tt : \implies \: \orange{ \underline \bold{7x = - 11y - 15}} \\ \\ \orange{ \underline \bold{divide \: both \: sides \: by \: 7}} \\ \\ \tt : \implies \:\orange{ \underline \bold{x = \frac{ - 11y \: - 15}{7} }} \\ itz \: shiwam.............$

Answer 2

Answer:

Solution of the pair of linear equations 3x - 4y + 11 = 0 and

                                                                    7x + 11y + 15 = 0 is

x = $\dfrac{-181}{61}$

y = $\dfrac{32}{61}$

Step-by-step explanation:

$\implies$ We can find the solutions of a pair linear equations by substitution or elimination method

$\implies$ Here I'm gonna use elimination method

$\implies$ In this method, we make the co-efficient of either x or y equal then cancel them out

$\implies$ 3x - 4y + 11 = 0     -------(1)

(1) * 11 $\implies$ 33x - 44y + 121 = 0    -------(3)

$\implies$ 7x + 11y + 15 = 0     -------(2)

(2) * 4 $\implies$ 28x + 44y + 60 = 0    -------(4)

(3) + (4) $\implies$ 33x - 44y + 121 + 28x + 44y + 60 = 0 + 0

$\implies$ 61x + 181 = 0

$\implies$ 61x = -181`

$\implies$ x = $\dfrac{-181}{61}$

Substituting this value in (1) we get,

$3 * \dfrac{-181}{61} - 4y + 11 = 0$

$\implies$  $4y = \dfrac{-543+11*61}{61}=\dfrac{671-543}{61}=\dfrac{128}{61}$

$\implies$  $y = \dfrac{128}{61 * 4}=\dfrac{32}{61}$