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5c=7(2c+3-12=
We move all terms to the left:
5c-(7(2c+3-12)=0
We add all the numbers together, and all the variables
5c-(7(2c-9)=0
We calculate terms in parentheses: -(7(2c-9), so:
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| 5c=7(2c+3-12 | | 10(z+1)=5(z-3)z= | | 5a*16=9 | | -(12-3c)=-66 | | 4m-8=2m+20 | | (1-y)^2-y^2=7 | | (1-y*y)-(y*y)=7 | | 1000/x+Y=0 | | 2x=-x+ | | 2(1-h)-3(h+2)=-1 | | -3x=5x+ | | 3.5*x=100 | | X-4y+1=0X= | | m=2(0,4) | | m=2;(0,4) | | m=2/(0,4) | | 1.5*x=0.8 | | 3=p+21 | | 11-5a=51a= | | 100*x=30 | | 3x+38/11=7 | | m=4-10 | | (600+x)/3*200=1000x | | x+19=3x+25 | | x^2-12x+39=7 | | h+5=11 | | 2x–7=11¨ | | 2x–7=11x= | | 2x-(8x+1)-(x+2)×5=9 | | m/4+5=7 | | 5j+4=29 | | h/2-6=3 |