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(7x)=(2x+27
We move all terms to the left:
(7x)-((2x+27)=0
We calculate terms in parentheses: -((2x+27), so:
(2x+27
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(2x+27
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(2x+27
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| q=28÷7 | | 2x/3+x=24 | | 2x+16+4-16/8=-2x | | 17+3=6+t | | -10v−8=10−8v | | j=85−10 | | 8n=7n−9 | | y=-7/2+86 | | x^2+3/2x−1/2=−1 | | z=84-14 | | -2t=6+4t | | x2+32x−12=−1 | | 24x-6=210 | | 1/5=x/525 | | 1/5=x/528 | | 64=-16t^2+0 | | 24x-6=114 | | -10+3g=10+7g | | 23=5+3c | | x^-1=(8/19) | | 8n^2(2n+7)+(2n+7)=0 | | 0=-16t^2-64 | | -8d+9=-5d | | 4x-6=2×+3 | | 41=4-p+5 | | -3(x-6)=-5+2 | | -6(3-5n)+4=-86 | | p=-5p+6 | | -7p-16(5p+14)=0 | | 2(x-3)^(2)-8=0 | | (36+x)/12=10 | | X+59x+51=84 |