This encounters adding, subtracting and finding the least usual multiple.
You are watching: P1v1/t1=p2v2/t2 solve for t2
Step by step Solution
Reformatting the intake :Changes made to your input need to not influence the solution: (1): "t2" was replaced by "t^2". 5 more similar replacement(s).
Rearrange:Rearrange the equation by individually what is come the appropriate of the equal authorize from both sides of the equation : p^1*v^1/t^1-(p^2*v^2/t^2)=0
Step by action solution :
Step 1 :
v2 leveling —— t2Equation in ~ the finish of action 1 : (v1) v2 ((p1)•————)-((p2)•——) = 0 (t1) t2
step 2 :v leveling — tEquation at the finish of action 2 : v p2v2 ((p1) • —) - ———— = 0 t t2
Step 3 :Calculating the Least common Multiple :3.1 uncover the Least typical Multiple The left denominator is : t The ideal denominator is : t2
Least usual Multiple: t2Calculating multiplier :
3.2 calculate multipliers for the 2 fractions signify the Least common Multiple by L.C.M signify the Left Multiplier by Left_M denote the best Multiplier through Right_M represent the Left Deniminator by L_Deno signify the best Multiplier through R_DenoLeft_M=L.C.M/L_Deno=tRight_M=L.C.M/R_Deno=1Making indistinguishable Fractions :
3.3 Rewrite the 2 fractions into equivalent fractionsTwo fountain are dubbed equivalent if they have actually the same numeric value. For instance : 1/2 and 2/4 are equivalent, y/(y+1)2 and also (y2+y)/(y+1)3 are indistinguishable as well. To calculation equivalent portion , main point the numerator of each fraction, through its respective Multiplier.
L. Mult. • L. Num. Pv • t —————————————————— = —————— L.C.M t2 R. Mult. • R. Num. P2v2 —————————————————— = ———— L.C.M t2 adding fractions that have actually a common denominator :3.4 including up the two tantamount fractions add the two indistinguishable fractions which now have actually a common denominatorCombine the molecule together, placed the amount or difference over the common denominator then mitigate to lowest state if possible:
pv • t - (p2v2) pvt - p2v2 ——————————————— = —————————— t2 t2
Step 4 :Pulling out favor terms :4.1 pull out like factors:pvt - p2v2=-pv•(pv - t)
Equation in ~ the end of action 4 : -pv • (pv - t) —————————————— = 0 t2
Step 5 :When a portion equals zero :5.1 once a fraction equals zero ...Where a fraction equals zero, its numerator, the component which is above the fraction line, have to equal zero.Now,to eliminate the denominator, Tiger multiplys both political parties of the equation through the denominator.Here"s how:
-pv•(pv-t) —————————— • t2 = 0 • t2 t2 Now, on the left hand side, the t2 cancels out the denominator, while, on the ideal hand side, zero time anything is tho zero.The equation now takes the shape:-pv • (pv-t)=0
Theory - root of a product :5.2 A product of several terms equates to zero.When a product of 2 or much more terms equates to zero, then at least one of the terms must be zero.We shall currently solve each term = 0 separatelyIn various other words, we room going to deal with as plenty of equations as there room terms in the productAny systems of hatchet = 0 solves product = 0 together well.
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Solving a single Variable Equation:5.3Solve-pv=0 setting any of the variables to zero solves the equation:p=0v=0Solving a solitary Variable Equation:
5.4Solvepv-t=0 In this form of equations, having an ext than one variable (unknown), you need to specify for which change you desire the equation solved.We shall not take care of this form of equations in ~ this time.