Question

3x - 5y -19 =0 and 7x+3y+1=0 solve this by using substitution method​

Answer 1

Ans : x=13/11 and y=−34/11

3x−5y−19=0;7x+3y+1=0

Step: Solve3x−5y−19=0 for x:

3x−5y−19=0

3x−5y−19+5y=0+5y(Add 5y to both sides)

3x−19=5y

3x−19+19=5y+19(Add 19 to both sides)

3x=5y+19

(Divide both sides by 3)

$\frac{3x}{3} = \frac{5y}{3} + \frac{19}{3}$

${x} = \frac{5y}{3} + \frac{19}{3}$

Now

substitute

$\frac{5y}{3} + \frac{19}{3}$

for x in 7x+3y+1=0:

$7( \: \frac{5y}{3} + \frac{19}{3}) +3y+1=0:$

$\frac{44y}{3} + \frac{136}{3} = 0$

$\: y = \frac{ - 34}{11}$

Now , substitute value of y in

${x} = \frac{5y}{3} + \frac{19}{3}$

We get ,

$x = \frac{13}{11}$

Thank you .

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