SOLUTION: I am doing the best i can with algebra can you check my problem and let me know what is wrong? 7(3x+4)=8(2x+5)+13 21x+28=8(2x+5)+13 21x+28=16x+40+13 -16x -16x 28-40+13

Question 68460: I am doing the best i can with algebra can you check my problem and let me know what is wrong?
7(3x+4)=8(2x+5)+13
21x+28=8(2x+5)+13
21x+28=16x+40+13
-16x -16x
28-40+13
28-53
=25
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
7(3x+4) = 8(2x+5) + 13
:
21x + 28 = 8(2x+5) + 13
:
21x + 28 = 16x + 40 + 13
:
21x - 16x + 28 = 16x -16x + 53; subtracted 16x from both sides, added 40 & 13
:
5x + 28 = 53
:
Subtract 28 from both sides and you have;
5x = 53 - 28
5x = 25
1x = 25/5; divided both sides by 5
x = 5
:
:
Check the solution in original equation:
7(3x+4) = 8(2x+5) + 13
7[3(5) + 4] = 8[2(5)+ 5] + 13
7(15 + 4) = 8(10 + 5) + 13
7(19) = 8(15) + 13
133 = 120 + 13 proves out solution

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