9514 1404 393
x = 39
Step-by-step explanation:
The usual approach is ...
12x -48 = 10x +30 . . . . use the distributive property to eliminate parentheses
2x -48 = 30 . . . . . . . . . subtract 10x from both sides [1]
2x = 78 . . . . . . . . . . . . add 48 to both sides [2]
x = 39 . . . . . . . . . . . . divide both sides by the coefficient of x
The solution is x = 39.
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Check
4(3x -12) = 5(2x +6) . . . . given
4(3·39 -12 = 5(2·39 +6) . . . . substitute for x
4(117 -12) = 5(78 +6) . . . . . . . multiply
4(105) = 5(84) . . . . . . . . . . . . add
420 = 420 . . . . . . . . . . . . . . multiply
The found value of x makes the equation true, so is the solution.
Additional Notes
[1] The point of the solution process is to put the variable on one side of the equal sign and a constant on the other side. To that end, we subtract the smallest variable term from both sides of the equation. The net result is a variable term on one side of the equal sign that has a positive coefficient.
We could have subtracted either of the constant terms first, and subtracted the variable term second. In this latter approach, the variable term subtracted is the one on the same side of the equation as the remaining constant term. Unless care is taken to select the constant on the side of the largest variable term (for subtraction), the net result may be a negative variable term. Technically, that works just as well, but can tend to increase the probability of an error being made.
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[2] We notice that all of the terms in the equation after the last step have the x-coefficient as a common factor. This gives us the opportunity to divide by that coefficient at this stage. Doing this would give x-24=15. Then adding 24 would give the same final solution. Sometimes I like to work equations this way so the variable is "bare" (has a coefficient of 1) sooner, and the numbers are smaller.
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In this discussion, when we talk about subtracting or dividing, we mean the entire equation is involved. The properties of equality tell us the equal sign remains valid if the same operation is performed on both sides of the equation. Our discussion here takes for granted that understanding. We have occasionally said "[do such and such] to both sides". Even where it isn't stated, the "to both sides" always applies.
