4 is a constant, so you can pull it out of the integral. Use a u substitution with u = 3-x and du = -dx. If it's a definite integral, remember to change the limits of integration. The integral is then simply 1/u which integrates to be ln u. Substitute back in 3-x for u and you have the answer to be: -4*ln (3-x) + C
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"2 over 3x equals 4" is an equation in a single variable, x.2/(3x) = 4 => 3x/2 = 1/4 => 3x = 2/4 = 1/2 => x = 1/6.
4=3x-14 18=3x I carried the 14 over and it became positive. x=6 I divided 18 by 3.
If: 3x/4 -5 = x/3 +10 Then: x = 36
3x * (x - 4) = 3x*x - 3x*4 = 3x^2 - 12x.
For any x ≠ 0, x2 -10/x2 - 4 +3x/x2 - 4 LCD = x2, multiply each term by their missing element of LCD = (x4 + 10 +3x - 8x2)/x = (x4 - 8x2 + 3x + 10)/x2