How do you solve # 8c-(c-5)c+17#?

Question

How do you solve # 8c-(c-5)>c+17#?

Hazel Anglin · Accepted Answer

#c>2#

distribute bracket on left side of inequality.

#8c-c+5>c+17#

#rArr7c+5>c+17#

collect terms in c on left side and numeric values on right side.

subtract c from both sides.

#7c-c+5>cancel(c)cancel(-c)+17#

#rArr6c+5>17#

subtract 5 from both sides.

#6c cancel(+5)cancel(-5)>17-5#

#rArr6c>12#

To solve for c, divide both sides by 6

#(cancel(6) c)/cancel(6)>12/6#

#rArrc>2" is the solution"#

Jameson Allen · Answer

undefined To solve the inequality 8c - (c - 5) > c + 17:

1. Distribute the negative sign inside the parentheses: 8c - c + 5 > c + 17.
2. Combine like terms: 7c + 5 > c + 17.
3. Subtract c from both sides to isolate the variable: 6c + 5 > 17.
4. Subtract 5 from both sides: 6c > 12.
5. Divide both sides by 6: c > 2.

Eva Adamson · Answer

undefined To solve the inequality $8c - (c - 5) > c + 17$, follow these steps:

1. Distribute the negative sign inside the parentheses:
   $$8c - c + 5 > c + 17$$

2. Combine like terms on both sides of the inequality:
   $$7c + 5 > c + 17$$

3. Subtract $c$ from both sides of the inequality to isolate the variable term:
   $$7c - c + 5 - c > c - c + 17$$
   $$6c + 5 > 17$$

4. Subtract 5 from both sides of the inequality:
   $$6c + 5 - 5 > 17 - 5$$
   $$6c > 12$$

5. Divide both sides by 6 to solve for $c$:
   $$\frac{6c}{6} > \frac{12}{6}$$
   $$c > 2$$

So, the solution to the inequality is $c > 2$.

How do you solve # 8c-(c-5)&gt;c+17#?

How do you solve # 8c-(c-5)>c+17#?