Question

solve equation 0.4x + 0.3y = 1.7 and 0.7x -0.2y=0.8 by substitution method

Answer 1

Answer: x=2,y=3

Step-by-step explanation: 0.4x+0.3y=1.7,  0.7x-0.2y=0.8

multiply both by 10 to get 4x+3y=17---eq 1

and 7x-2y=8---eq 2

so,from eq2, x=8+2y/7

substitute value of y in eq1 to get

4(8+2y/7)+3y=17

32+8y/7+3y=17

32+8y+21y/7=17..(taking LCM=7)

32+29y=17.7

29y=119-32

29y=87

==> *y=3*

substitute value of y in eq2 to get

7x-2(3)=8

7x-6=8

7x=8+6

==> *x=2*

hope this helps you...:))

Answer 2

Answer:

x=2,y=3

Step-by-step explanation:

0.4x+0.3y=1.7,  0.7x-0.2y=0.8

multiply both by 10 to get 4x+3y=17--- 1

and 7x-2y=8---2

so,from 2, x=8+2y/7

substitute value of y in 1 to get

4(8+2y/7)+3y=17

32+8y/7+3y=17

32+8y+21y/7=17

32+29y=17.7

29y=119-32

29y=87

y=3

substitute value of y in eq2 to get

7x-2(3)=8

7x-6=8

7x=8+6

x=2