# Solve for c and d: 3c – d = 30 and 5c – 3d = 10

Jyotsana | Student, Grade 10 | (Level 1) Valedictorian

Posted on

3c-d=30

3c-d=30=3c

-d=30-3c

-d/-1=30/-1-3c/-1

d=-30+3c

5c-3(-30+3c)=10

5c+90-9c=10

-4c+90=10

-4c+90-90=10-90

-4c=-80

-4c/-4=-80/-4

c=20

3(20)-d=30

60-d=30

60-d-60=30-60

-d=-30

-d/-1=-30/-1

d=30

bloo786 | Student, Grade 10 | (Level 1) eNoter

Posted on

3c-d=30    (i)                             5c-3d=10   (ii)

3c-30=d

put the value in the second equ

5c-3(3c-30)=10

5c-9c+90=10

4c=80

c=80/4

c=20

now we have the value of c

we can put this value in any equ to find the value of d or we can do this the long way

3c-d=30

3c=30+d

c=30/3+d/3

c=10+d/3

puting this value in the second equ

5(10+d/3)-3d=10

50+5d/3-3d=10

50-10=-5d/3+3d

40=-5d/3+9d/3              (took L.C.M on the right side of equals to)

40=4d/3

40x3=4d

120=4d

d=30

now we have the value of d.

we can put the value of c and d in any equation to check the answer.

(NOTE: This is the substution method for the solution for simultaneous linear equations. We can also use the comparison method and many more.)

penemra | Student, Grade 9 | (Level 1) Valedictorian

Posted on

3c-d=30 (-firstX3)

5c-3d=10(-wecondx1)

or 9c-3d=90

5c-3d=10

-   +     -

_____________

4c=80

c=80/4 = 20

5c-3d=10

100-3d=10

-3d=-90

d=-90/-3 = 30

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll re-write the first equation:

3c - 30 = d (1)

We'll replace (1) in the 2nd equation:

5c - 3(3c - 30 ) = 10

We'll remove the brackets and we'll move all terms to the left side:

5c - 9c + 90 - 10 = 0

We'll combine like terms:

-4c + 80 = 0

-4c = -80 => c = 20

d = 60 - 30 => d = 30