Question

please solve this equation ​

Answer 1

Answer:

3-2(3x+4y)=x

3-6x-8y=x

3-7x-8y=0

Hence 3-8y=7x

therefore ,x=(3-8y)/7 --- eqn1

(x-3)/4-(y-4)/5=2*1/10

Here take LCM on LHS and Simplify RHS

5(x-3)-4(y-4)/20=1/5

(5x-15-4y+16)/20=1/5

now cancel 5 and 20

(5x-4y+1)/4=1

take denominator to RHS

5x-4y+1=4

5x-4y=3 ---eqn2

now put eqn 1 in the eqn2

Means substitute value of x in eqn 2

5((3-8y)/7)-4y=3

(15-40y)/7-4y =3

15-40y-28y=21

-68y=6

y=-3/34

now put value of y in eqn 1

(3-8y)/7=X

but y=-3/34

Therefore

(3-8*(-3/34))/7=x

(3+(24/34))/7

(3*34+24)/34*7

(102+24)/(34*7)

126/238

x=63/119

Hence x=63/119

y=-3/34