# Whats the answer to 24-2c+3=2c+3+c

## Answers

Answer: c = [tex]\frac{24}{5}[/tex]Step-by-step explanation:Hey there!Step 1: Cancel 33 on both sides.24 - 2c = 2c + cStep 2: Simplify 2c + c to 3c.24 - 2c = 3cStep 3: Add 2c to both sides.24 = 3c + 2cStep 4: Simplify 3c + 2c to 5c.24 = 5cStep 5: Divide both sides by 5[tex]\frac{24}{5}[/tex] = cStep 6: Switch Sides.c = [tex]\frac{24}5}[/tex]~I hope I helped you! :)~

Answer: c≓−2.135980077How to solve it: Step by step solution :STEP1: 2c + 5 Simplify —————— c Equation at the end of step1: 2 (2c+5) (c-((—•(c2))•(2c-5)))-(((3•(c3))•——————)+2) = 0 6 c STEP 2 :Equation at the end of step2: 2 (2c+5) (c-((—•(c2))•(2c-5)))-((3c3•——————)+2) = 0 6 c STEP3:Dividing exponential expressions 3.1 c3 divided by c1 = c(3 - 1) = c2Equation at the end of step3: 2 (c-((—•(c2))•(2c-5)))-(3c2•(2c+5)+2) = 0 6 STEP 4 : 1 Simplify — 3Equation at the end of step4: 1 (c - ((— • c2) • (2c - 5))) - (6c3 + 15c2 + 2) = 0 3 STEP5:Equation at the end of step 5 c2 (c - (—— • (2c - 5))) - (6c3 + 15c2 + 2) = 0 3 STEP6:Equation at the end of step 6 c2 • (2c - 5) (c - —————————————) - (6c3 + 15c2 + 2) = 0 3 STEP7:Rewriting the whole as an Equivalent Fraction 7.1 Subtracting a fraction from a wholeRewrite the whole as a fraction using 3 as the denominator : c c • 3 c = — = ————— 1 3 Equivalent fraction : The fraction thus generated looks different but has the same value as the wholeCommon denominator : The equivalent fraction and the other fraction involved in the calculation share the same denominatorAdding fractions that have a common denominator : 7.2 Adding up the two equivalent fractionsAdd the two equivalent fractions which now have a common denominatorCombine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible: c • 3 - (c2 • (2c-5)) -2c3 + 5c2 + 3c ————————————————————— = ——————————————— 3 3 Equation at the end of step7: (-2c3 + 5c2 + 3c) ————————————————— - (6c3 + 15c2 + 2) = 0 3 STEP8:Rewriting the whole as an Equivalent Fraction 8.1 Subtracting a whole from a fractionRewrite the whole as a fraction using 3 as the denominator : 6c3 + 15c2 + 2 (6c3 + 15c2 + 2) • 3 6c3 + 15c2 + 2 = —————————————— = ———————————————————— 1 3 STEP9:Pulling out like terms 9.1 Pull out like factors : -2c3 + 5c2 + 3c = -c • (2c2 - 5c - 3) Trying to factor by splitting the middle term 9.2 Factoring 2c2 - 5c - 3 The first term is, 2c2 its coefficient is 2 .The middle term is, -5c its coefficient is -5 .The last term, "the constant", is -3 Adding up the two equivalent fractions -c • (c-3) • (2c+1) - ((6c3+15c2+2) • 3) -20c3 - 40c2 + 3c - 6 ———————————————————————————————————————— = ————————————————————— 3 3 STEP10:Pulling out like terms 10.1 Pull out like factors : -20c3 - 40c2 + 3c - 6 = -1 • (20c3 + 40c2 - 3c + 6) Checking for a perfect cube : 10.2 20c3 + 40c2 - 3c + 6 is not a perfect cubeSolving 2c2-5c-3 = 0 by the Quadratic Formula . According to the Quadratic Formula, c , the solution for Ac2+Bc+C = 0 , where A, B and C are numbers, often called coefficients, is given by : - B ± √ B2-4AC c = ———————— 2A In our case, A = 2 B = -5 C = -3Accordingly, B2 - 4AC = 25 - (-24) = 49Applying the quadratic formula : 5 ± √ 49 c = ————— 4Can √ 49 be simplified ?Yes! The prime factorization of 49 is 7•7 To be able to remove something from under the radical, there have to be 2 instances of it (because we are taking a square i.e. second root).√ 49 = √ 7•7 = ± 7 • √ 1 = ± 7So now we are looking at: c = ( 5 ± 7) / 4Two real solutions:c =(5+√49)/4=(5+7)/4= 3.000or:c =(5-√49)/4=(5-7)/4= -0.500One solution was found : c ≓ -2.135980077