Question

3x-5y-4=0 and 9x=2y+7 solve by elimination and substituting method

Answer 1

.... ..... ....... ........ .......... ............ ............ ..............

Answer 2

Answer:

$x=\frac{9}{13}$  and $y=-\frac{5}{13}$  

Step-by-step explanation:

Given : $3x-5y-4=0$

            $9x=2y+7$

To Find :solve by elimination and substituting method

Solution :

Substitution method

$3x-5y-4=0$ ---A

$9x=2y+7$   ---B

Substitute the value of x from A in B

$9(\frac{4+5y}{3})=2y+7$  

$3(4+5y)=2y+7$  

$12+15y=2y+7$  

$13y=-5$  

$y=-\frac{5}{13}$  

Substitute the value of y in A to get value of x

$3x-5(-\frac{5}{13})-4=0$

$3x+\frac{25}{13})-4=0$

$39x+25=52$

$39x=27$

$x=\frac{27}{39}$

$x=\frac{9}{13}$

So, $x=\frac{9}{13}$  and $y=-\frac{5}{13}$  

Elimination method

$3x-5y=4$ ---A

$9x-2y=7$   ---B

Multiply A with 9 and B with 3

$27x-45y=36$ ---C

$27x-6y=21$   ---D

Now subtract D from C

$27x-45y-27x+6y=36-21$

$-45y+6y=36-21$

$-39y=15$

$y=\frac{-15}{39}$

$y=\frac{-5}{13}$

Substitute the value of y in A

$3x-5(-\frac{5}{13})-4=0$

$3x+\frac{25}{13}-4=0$

$39x+25=52$

$39x=27$

$x=\frac{27}{39}$

$x=\frac{9}{13}$

So, $x=\frac{9}{13}$  and $y=-\frac{5}{13}$