how do you solve the problem? 1/3(3a-1)+3/5(4a+3)=3/5(a-7)+1

Question

My question is about solving linear equations involving fractions.

Answer 1

⅓(3a-1)+⅗(4a+3)=⅗(a-7)+1.

Find the LCM of 3 and 5, which is 3×5=15. Multiply through by 15:

5(3a-1)+9(4a+3)=9(a-7)+15,

15a-5+36a+27=9a-63+15,

51a+22=9a-48,

51a-9a=-48-22,

42a=-70,

a=-70/42=-5/3, so the answer is a=-5/3.

CHECK

LHS: ⅓(3a-1)+⅗(4a+3)=⅓(-5-1)+⅗(-20/3+3)=⅓(-6)+⅗(-11/3)=-2-11/5=-21/5;

RHS=⅗(a-7)+1=⅗(-5/3-7)+1=⅗(-26/3)+1=-26/5+1=-21/5.

So LHS=RHS, confirming solution a=-5/3.