Question

Q.4. Solve the following [pair of equations by the elimination method:
(a) 2x – 3y =7 ; 5x + 2y = 10

Answer 1

Answer:

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Answer 2

ANSWER↪➡➡x=44/19 and y=15/-19

☢☢QUESTION➡Solve the following [pair of equations by the elimination method:

(a) 2x – 3y =7 ; 5x + 2y = 10

☢☢TO FIND➡find the value of

'x' and'y'

↪using elimination method...

$\bf\ 2x - 3y = 7 \purple{ - - - - - - - (1)}\\ \\ \bf\ 5x + 2y = 10 \green{ - - - - - - - (2)}\\ \\ (1) \times 2\bf\red{ \implies: } 2(2x - 3y) = 7 \times 2\\ \\ \bf\red{ \implies: }4x - 6y = 14 \orange{- - - - - - (3)} \\ \\ \bf\ (2) \times 3 \bf\red{ \implies:}3(5x + 2y) = 10 \times 3 \\ \\ \bf\red{ \implies:}15x + 6y = 30 \red{- - - - - - - - (4)} \\ \\ \underline{now } \\ \\eqn(4) + eqn(3)\bf\red{ \implies:} \\ \\ \bf\red{ \implies:}(15x + 6y) + (4x - 6y) = 30 + 14 \\ \\ \bf\red{ \implies:}15x \cancel{+ 6y} + 4x \cancel{- 6y }= 44 \\ \\ \bf\red{ \implies:}19x = 44 \\ \\ \bf\red{ \implies:} \boxed{x = \frac{44}{19}} \\ \\ \underline{value \: of \: x \: put \: in \: (1)} \\ \\ \bf\ 2 \times \frac{44}{19} - 3y = 7 \\ \\ \bf\red{ \implies:} - 3y = 7 - \frac{88}{19} \\ \\ \bf\red{ \implies:} - 3y = \frac{133 - 88}{19} \\ \\ \bf\red{ \implies:} - 3y = \frac{45}{19} \\ \\ \bf\red{ \implies:}y = \frac{45}{19 \times - 3} \\ \\ \bf\red{ \implies:} \boxed{y = \frac{15}{ - 19} } \\ \\ \large\boxed{ \fcolorbox{red}{purple}{hope \: it \: help \: you}}$