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grandymaker [24]
3 years ago
12

X/2 + 1 = 1/4x - 6 solve for x

Mathematics
2 answers:
malfutka [58]3 years ago
6 0

x=-28......................

Natalka [10]3 years ago
5 0
<h3><em>Answer: x = -28</em></h3><h3 /><h3><em>Step-by-step explanation:</em></h3><h3><em>Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation : </em></h3><h3><em>x/2+1-(1/4*x-6)=0 </em></h3><h3><em>                     1 </em></h3><h3><em>Simplify      — </em></h3><h3><em>                    4</em></h3><h3 /><h3><em>    x                         1          </em></h3><h3><em>  (— +  1)    -        ((— • x) -  6)  = 0  </em></h3><h3><em>   2                        4         </em></h3><h3 /><h3><em>Subtracting a whole from a fraction </em></h3><h3 /><h3><em>Rewrite the whole as a fraction using  4  as the denominator : </em></h3><h3 /><h3><em>            6             6 • 4 </em></h3><h3><em>    6 =  —  =  ————— </em></h3><h3><em>           1             4   </em></h3><h3><em>Equivalent fraction : The fraction thus generated looks different but has the same value as the whole </em></h3><h3 /><h3><em>Common denominator : The equivalent fraction and the other fraction involved in the calculation share the same denominator</em></h3><h3 /><h3 /><h3><em> Adding up the two equivalent fractions </em></h3><h3><em>Add the two equivalent fractions which now have a common denominator </em></h3><h3 /><h3><em>Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible: </em></h3><h3 /><h3><em> x - (6 • 4)                               x - 24 </em></h3><h3><em> ———————————  =  —————— </em></h3><h3><em>      4                                            4   </em></h3><h3><em>   </em></h3><h3 /><h3><em>    x                    (x - 24) </em></h3><h3><em>  (— +  1) -  ————————  = 0  </em></h3><h3><em>   2                 4    </em></h3><h3 /><h3><em>                  x </em></h3><h3><em> Simplify   — </em></h3><h3><em>                  2</em></h3><h3 /><h3 /><h3><em>    x                  (x - 24) </em></h3><h3><em>  (— +  1) -  ————————  = 0  </em></h3><h3><em>   2                 4    </em></h3><h3 /><h3><em>Adding a whole to a fraction </em></h3><h3 /><h3><em>Rewrite the whole as a fraction using  2  as the denominator : </em></h3><h3 /><h3><em>           1            1 • 2 </em></h3><h3><em>    1 =  —  =  ————— </em></h3><h3><em>          1              2  </em></h3><h3 /><h3><em> Adding up the two equivalent fractions </em></h3><h3 /><h3><em>   x + 2              x + 2 </em></h3><h3><em> —————  =  ————— </em></h3><h3><em>    2                    2  </em></h3><h3 /><h3><em>          (x + 2)            (x - 24) </em></h3><h3><em>  ——————— -  ————————  = 0  </em></h3><h3><em>     2                               4    </em></h3><h3 /><h3><em>Find the Least Common Multiple </em></h3><h3 /><h3><em>      The left denominator is :       2  </em></h3><h3 /><h3><em>      The right denominator is :       4  </em></h3><h3 /><h3><em>        Number of times each prime factor </em></h3><h3><em>        appears in the factorization of: </em></h3><h3><em> Prime  </em></h3><h3><em> Factor   Left  </em></h3><h3><em> Denominator   Right  </em></h3><h3><em> Denominator   L.C.M = Max  </em></h3><h3><em> {Left, Right}  </em></h3><h3><em>2 1 2 2 </em></h3><h3><em> Product of all  </em></h3><h3><em> Prime Factors  2 4 4 </em></h3><h3 /><h3><em>      Least Common Multiple: </em></h3><h3><em>      4 </em></h3><h3 /><h3 /><h3><em>Calculate multipliers for the two fractions </em></h3><h3 /><h3 /><h3><em>    Denote the Least Common Multiple by  L.C.M  </em></h3><h3><em>    Denote the Left Multiplier by  Left_M  </em></h3><h3><em>    Denote the Right Multiplier by  Right_M  </em></h3><h3><em>    Denote the Left Deniminator by  L_Deno  </em></h3><h3><em>    Denote the Right Multiplier by  R_ Deno  </em></h3><h3 /><h3><em>   Left_ M = L.C.M / L_ Deno = 2 </em></h3><h3 /><h3><em>   Right_ M = L.C.M / R_Deno = 1</em></h3><h3 /><h3><em> Rewrite the two fractions into equivalent fractions </em></h3><h3 /><h3><em>Two fractions are called equivalent if they have the same numeric value. </em></h3><h3 /><h3><em>For example :  1/2   and  2/4  are equivalent,  y/(y+1)2   and  (y2+y)/(y+1)3  are equivalent as well. </em></h3><h3 /><h3><em>To calculate equivalent fraction , multiply the Numerator of each fraction, by its respective Multiplier. </em></h3><h3 /><h3><em>   L. Mult. • L. Num.      (x+2) • 2 </em></h3><h3><em>   ——————————————————  =   ————————— </em></h3><h3><em>         L.C.M                 4     </em></h3><h3 /><h3><em>   R. Mult. •  R. Num.                                   ( x-24) </em></h3><h3><em>   ——————————————————  =   —————— </em></h3><h3><em>                       L.C.M                                            4   </em></h3><h3 /><h3><em>Adding up the two equivalent fractions </em></h3><h3 /><h3><em> (x+2) • 2 - ((x-24))                                           x + 28 </em></h3><h3><em> ————————————————————  =  —————— </em></h3><h3><em>          4                                                                  4   </em></h3><h3><em> </em></h3><h3><em>      x + 28 </em></h3><h3><em>  ——————  = 0  </em></h3><h3><em>      4   </em></h3><h3 /><h3 /><h3><em>When a fraction equals zero ... </em></h3><h3><em>Where a fraction equals zero, its numerator, the part which is above the fraction line, must equal zero. </em></h3><h3 /><h3><em>Now, to get rid of the denominator, Tiger multiplys both sides of the equation by the denominator. </em></h3><h3 /><h3><em>Here's how: </em></h3><h3 /><h3><em>   x+28 </em></h3><h3><em>  ———— • 4 = 0 • 4 </em></h3><h3><em>    4   </em></h3><h3><em>Now, on the left hand side, the  4  cancels out the denominator, while, on the right hand side, zero times anything is still zero. </em></h3><h3 /><h3><em>The equation now takes the shape : </em></h3><h3><em>   x+28  = 0</em></h3><h3 /><h3><em>Solve  :  x+28 = 0  </em></h3><h3 /><h3><em> Subtract  28  from both sides of the equation :  </em></h3><h3><em>                      x = -28</em></h3><h3 /><h3 /><h3 /><h3 /><h3>I hope it is helpful?</h3>
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The probability that a student has a Visa card (event V) is .63. The probability that a student has a MasterCard (event M) is .1
Aleksandr-060686 [28]

Answer:

a)  The probability that a student has either a Visa card or a MasterCard is 0.71.  

b) V and M are not independent.

Step-by-step explanation:

Given : The probability that a student has a Visa card (event V) is 0.63. The probability that a student has a MasterCard (event M) is 0.11. The probability that a student has both cards is 0.03.

To find :

a) The probability that a student has either a Visa card or a MasterCard ?

b)  In this problem, are V and M independent ?

Solution :

The probability that a student has a visa card(event V) is P(V)= 0.63

The probability that a student has a MasterCard (event M) is P(M)= 0.11

The probability that a student has both cards  is P(V \cap M)=0.03

a) Probability that a student has either a Visa card or a Master Card is given by,

P(V \cup M) = P(V) + P(M) - P(V\cap M)

P(V \cup M) = 0.63+ 0.11- 0.03

P(V \cup M) =0.74- 0.03

P(V \cup M) =0.71

The probability that a student has either a Visa card or a MasterCard is 0.71.

b) Two events, A and B, are independent if P(A\cap B)=P(A)P(B)

For V and M to be independent the condition is satisfied,

P(V\cap M)=P(V)P(M)

Substitute the values,

0.03=0.63\times 0.11

0.03\neq 0.0693

So, V and M are not independent.

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