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

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Question:

whats the answer to 24-2c+3=2c+3+c

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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

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