Question

Solving simultaneous equations by algebraic methods
3c-5d=17
4c+3d=13 please show working

Answer 1

I wouldn't use subsitution because that could be messy

I'll use elimination


multiply first equation by 3 and 2nd by 5 and add them

9c-15d=51
20c+15d=65 +
29c+0d=116

29c=116
c=4

sub back

3c-5d=17
3(4)-5d=17
12-5d=17
-5d=5
d=-1



c=4
d=-1

Answer 2

4(3c-5d=17)-3(4c+3d=13)

12c-20d-12c-9d=68-39

-29d=29

d=-1, now you can use either original equation to solve for c

3c-5(-1)=17

3c+5=17

3c=12

c=4

So c=4 and d= -1