The equalisation method requires us to eliminate one variable. We can do that in this case in two different ways.
If we make x the subject of each of these two equations we get:
3x=7y+24, and 5x=21-4y, so
x=(7y+24)/3, and x=(21-4y)/5.
Since x is equal to both expressions involving y, the two expressions must be equal:
(7y+24)/3=(21-4y)/5.
Now we cross-multiply to get rid of the fractions:
5(7y+24)=3(21-4y).
Now expand the parentheses by distribution:
35y+120=63-12y, collect the variable terms and constant terms together:
35y+12y=63-120,
47y=-57, so y=-57/47. By substitution we can find x=(21-4y)/5=(21+228/47)/5=1215/(47×5)=243/47.
So the solution is x=243/47 and y=-57/47. What a horrible answer, but it seems to be correct if the given equations are correct.
Another way is to get one of the variables to be the same in each equation by multiplying each equation by a constant. Let’s multiply the first equation by 4: 12x-96=28y; and the second equation by 7: 28y=147-35x. Both equations have a value 28y so we can equate the two expressions in x:
12x-96=147-35x, 12x+35x=147+96, 47x=243, x=243/47. And 3(243/47)-24=7y, 729/47-24=7y, (729-1128)/47=7y, -399/47=7y, -57/47=y. This is the same answer as before.