Solve for c 28-(3c+4)=2(c+6)+c

Math
28-(3c+4)=2(c+6)+c
Since c is on the right side of the equation, switch the sides so it is on the left side of the equation.
2(c+6)+c=28-(3c+4)
Simplify 2(c+6)+c.
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Simplify each term.
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Apply the distributive property.
2c+2⋅6+c=28-(3c+4)
Multiply 2 by 6.
2c+12+c=28-(3c+4)
2c+12+c=28-(3c+4)
Add 2c and c.
3c+12=28-(3c+4)
3c+12=28-(3c+4)
Simplify 28-(3c+4).
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Simplify each term.
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Apply the distributive property.
3c+12=28-(3c)-1⋅4
Multiply 3 by -1.
3c+12=28-3c-1⋅4
Multiply -1 by 4.
3c+12=28-3c-4
3c+12=28-3c-4
Subtract 4 from 28.
3c+12=-3c+24
3c+12=-3c+24
Move all terms containing c to the left side of the equation.
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Add 3c to both sides of the equation.
3c+12+3c=24
Add 3c and 3c.
6c+12=24
6c+12=24
Move all terms not containing c to the right side of the equation.
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Subtract 12 from both sides of the equation.
6c=24-12
Subtract 12 from 24.
6c=12
6c=12
Divide each term by 6 and simplify.
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Divide each term in 6c=12 by 6.
6c6=126
Cancel the common factor of 6.
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Cancel the common factor.
6c6=126
Divide c by 1.
c=126
c=126
Divide 12 by 6.
c=2
c=2
Solve for c 28-(3c+4)=2(c+6)+c
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