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Maths Aryabhatta

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Solve the following pair of linear equations using substitution method :-

2x+3y + 5 = 0 ; 5x-3y+9=0​

Sagot :

Answer:

(- 2, - [tex]\frac{1}{3}[/tex] )

Step-by-step explanation:

Given the 2 equations

2x + 3y + 5 = 0 → (1)

5x - 3y + 9 = 0 → (2)

Rearrange (1) expressing 3y in terms of x by subtracting 2x + 5 from both sides

3y = - 2x - 5

Substitute 3y = - 2x - 5 into (2)

5x - (- 2x - 5) + 9 = 0 ← distribute parenthesis on left side and simplify

5x + 2x + 5 + 9 = 0

7x + 14 = 0 ( subtract 14 from both sides )

7x = - 14 ( divide both sides by 7 )

x = - 2

Substitute x = - 2 into either of the 2 equations and solve for y

Substituting into (1)

2(- 2) + 3y + 5 = 0

- 4 + 3y + 5 = 0

1 + 3y = 0 ( subtract 1 from both sides )

3y = - 1 ( divide both sides by 3 )

y = - [tex]\frac{1}{3}[/tex]

solution is (- 2, - [tex]\frac{1}{3}[/tex] )

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