Question

What is the solution of 3x+y+1=0, 2x-3y+8=0 *

1 point

A) x= -1, y=2

B) x=1, y=2

C) x=2, y=3

D) x=4, y=5

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Answer 1

First, we will write the equations in their standard form.

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3x + y + 1 = 0

Transposing 1 to R.H.S. will give,

3x + y = –1

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2x – 3y + 8 = 0

Transposing 8 to R.H.S. will give,

2x – 3y = – 8

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Now we have the equations in their standard form.

3x + y = –1

2x – 3y = –8

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Using elimination method to find the value of x and y.

Let's eliminate y.

We will multiply the first eq. by 3 so that the coefficients of y in both the eq. becomes same.

3(3x+y= –1) = 9x+3y=–3

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Now we will add both the equations.

9x + 3y = –3

+(2x – 3y = –8)

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11x= –11

x = – 1

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Substituting the value of x in 2nd eq.

⇒2x – 3y = –8

⇒2(–1) – 3y = –8

⇒–2 – 3y = –8

⇒ 2 + 3y = 8

⇒3y= 8 – 2

⇒ 3y = 6

⇒ y = 6/3

⇒ y = 2

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x = – 1

y = 2

Therefore, option (A) is correct!