Solve for x. State any restrictions on the variables: a(x+c)=b(x-c)?
a) ax+ac=bx-bc ac+bc = bx-ax --> (b-a)·x=(a+b)·c *If b-a≠0 --> x= (a+b)·c / (b-a) *If b-a=0 and a+b=0 or c=0--> x any real number *If b-a=0 and a+b≠0 and c≠0 --> There aren't solutions b) (1/b-1/a)x = 1 --> (a-b)/(a·b) ·x = 1 with a,b≠0 *If a-b≠0 --> x=ab/(a-b) *If a-b=0 --> There aren't any solution saludos.
Solve each equation for x. find any restrictions.?
1) ax+bx=c ax + bx = c x*(a + b) = c x = c/(a + b) Restriction: a != -b 2) x-2/2=m+n (x-2)/2 = m+n (x-2) = 2(m+n) x = 2(m+n) + 2 No restrictions. 3) x/a+b= c (x/a) + b = c (x/a) = c - b x = a * (c - b) In the original equation, a != 0. If this problem was intended to be: x / (a+b) = c Then it reduces as: x = c * (a+b) with the restriction that a != -b. .
Solve for n, first stating any restrictions?
The restrictions are the numbers that will make the denominator zero. In this case, if n + 2 = 0, then n≠-2 is one restriction. If n - 5 = 0, then n≠5 is another restriction. Finally, n² - 3n - 10 = 0 will give the same two restrictions as the previous ones already mentioned. Multiply by the LCD: (n + 2)(n - 5) on both sides of the equation. This will result in the n + 2's canceling in the first part, the n - 5's canceling in the second part, and both the n + 2's and n -5's canceling in the third part. Your resulting equation is then: 3(n - 5) + 4(n + 2) = n - 19 Distribute: 3n - 15 + 4n + 8 = n - 19 Simplify: 7n - 7 = n - 19 Solve for n: 7n - n = -19 + 7 6n = -12 n = -2 Now, check this against your restrictions. Since n cannot equal -2, this problem has NO SOLUTION.
Solve for x. State any restrictions on the variable? Algebra help!?
4. 4/(9(x + 3)) = g 9(x + 3) = 4/g x + 3 = 4/(9g) x = 4/(9g) - 3 In the original equation, you do not want 0 to be your denominator because divided by 0 is undefined, so, by setting x + 3 = 0 implies x cannot be -3 is the restriction for the variable x. 5. a(x + c) = b( x - c) ax +ac = bx - bc ax - bx = -ac - bc x(a - b) = -c(a + b) x = -c(a + b)/(a - b) Again, here you do not want the denominator of your original equation to be 0, so the restriction is a ≠ b. 6. x + 3/t = t² x = t² - 3/t Same fashion here. t≠0 in the original equation. Hope it helps.
6 algebra II questions?
1) ax + bx -1 = 8 add 1 to both sides: ax + bx = 9 take out the x on the left: x(a+b) = 9 divide both sides by a + b: x = 9 / (a + b) restrictions would be: all real numbers such that a + b does not equal zero 2) as you know rectangles have 4 sides with opposite sides being the same length. Set up a rectangle with the side lengths of 5 and 7. Find the perimeter of that rectangle: 5 + 5 + 7 + 7 = 24 set up a ratio with the given lengths. (you're given that the actual rectangle is 96) so you say 96 is to 24 as n is to 5 and as m is to 7: 96:24 n:5 m: now solve for n and m: multiple across and then divide. (5 times 96 divided by 24) = 20 ( 7 times 96 divided by 24) = 28 Now you know the actual lengths of the sides are 20cm and 28 cm find the area: multiply length times width: 20 x 28 = 560 cm. 3) the wording of this problem is off. but it'd be something like 25 + 7a = 180 - a then you'd solve. I'm just confused of which angle you're meaning. 4) plug in all values for their variables (25 - 10) ^2 + 10(11)(18) simplyify: (15)^2 +1980 square 15: 225 + 1980 add: 2205 5) T 6) -9y - 7 = -13 -2y add 7 to both sides: -9y = -6 - 2y add 2y to both sides: -7y = -6 divide both sides by -7: y = -6 / -7 = 6/7
How do you solve ax-3x+5=a+b for x? it also wants me to state any restrictions on the variables.?
It's a variable comparison question ax-3x+5=a+b a(x-1)-3x+5=b a(x-1)+5=b+3x ax-a+5=b+3x now comparing variables on both side of the equation we get a=3 n a+b=5 therefore,b=2
How do I solve cryptarithmetic problems like BASE+BALL=GAMES?
Answer is:A=4,B=7,S=8,E=3,L=5,G=1,andM=9Explanation:Here we are provided with the following information BASE +BALL ------------ GAMESThere are seven distinct digits from 10 preliminary digits that are from [0-9]:A,B,S,E,LG,MAs we are just adding 2 numbers so possible carry overs are either 1 or 0But, when we are adding B+B it gives us some carry that is greater than zero as both 4 digit numbers add up to form a five digit number Hence, G can't be zeroSo, G=1Now our question will look somewhat like this: BASE +BALL ------------ 1 AMESThen,if 2 B's are added and they are giving us value greater than equal to 10Then,B must be greater than equal to 5Consider B=9 firstif B=9 then A=8 which is not possible as then B+B+1 is not equal to Aif B=8 then A=6 which is not possible for same reasonTake B=7 then A=4 which is possible as it is not leaving any carry over.So,M=A+A+x where x is carry over from S+L+y where y is carry over from E+Lknowing the fact that x and y can take maximum value of 1 and minimum value of 0We can sayM should be either 8 or 98 when x=09 when x=1Assuming M=8 that is x=0S+L+y=EE+L= 10*y+SS+L+y+L=10*y+S2*L=9*ywhere y can take either 0 or 1if y=0 then L=0 then S and E are not distint which is not possible according to question.if y=1then L=4.5 which is again not possible as L must be a single digit among 0 to 9So, M=8 is not possibleSo,our question further reduces to 1 74SE +74LL ------------ 1 49ESNow, S+L=10*1+E=10+E,andE+L=10*y+SSo,here y can take maximum and minimum values of 1 and 0 respectively.So,if y=0Then, E+L=S,or S-L=E,andS+L=10+EAdding both we get,2*S=2*E+10or S=E+5Subtracting 1st equation from 2nd we get2*L=10or L=5So,our question further reduces to 1 0 74SE +7455 ------------ 1 49ESSo,E+5=S,andS+5+0=E+10*1S+5=E+10S=E+5Now we have to choose digits from 0,1,2,3,4,5,6,7,8,and 9 other than 1,4,7,9,and 5.So,we can take values of E and S from 0,2,3,6,8 such that S=E+50 is there 5 can't be chosen2 is there 7 can't be chosen6 is there 11 can't be chosen8 is there 13 can't be chosen So, only possible pair left is 3 and 8 which satisfy our constraints.Hence,S=8 and E=3 1 0 7483 +7455 ------------ 1 4938So,A=4,B=7,S=8,E=3,L=5,G=1,andM=9
What's the difference between a basic solution, a feasible solution and a basic feasible solution in linear programming?
There are three stages of a linear programmingInitialization .Iteration .Termination.In Initialization phase we give a solution to the simplex matrix which moves from corner to corner in bounded region.Like when we give (0,0) as solution of simplex matrix it will move in either x direction or y direction depends upon rate of change of each other in objective function.e.g if in this case objective function is [math]5x+3y[/math] (maximize).Then in this case simplex will start from (0,0) and move along X direction and choose corner points till the constrained satisfied.This solution is called basic feasible solution.n variablesm constraintsAs per simplex method in linear programming all the constraints should be in equality form . So, all inequality will be converted into equality by using slack variables and artificial variables . So it is logically proven that [math]n>=m[/math]We can solve for m variables as there are m constraints. [math]n-m [/math]variables are considered as non basic variableAt termination stage there will be three possibilities as followsFeasible solutionUnbounded solutionInfeasible solutionIf it has feasible solution means it can satisfied all the constraints and lead to an optimal solution(might me more optimal also).Feasible solution means set which contains all the possible solution which follow all the constraints.Basic solution - Solution of a problem which satisfy all the condition. Basically in linear programming more importance is about basic feasible solution rather than basic solution.Edit 1-If m sources and n destination of the demand then [math]summation of resource = meeting demand at destination [/math]will be your constraint. Although it depends upon whether your problem is balanced or unbalanced.Feasible solution- A set of non negative individual allocation which satisfy all the given constraints is termed as feasible solution.Basic feasible solution(B.F.S) - For [math]m*n [/math]transportation problem if number of allocations are [math]m+n-1 [/math]ten it is called as basic feasible solution.Non degenerate B.F.S- when all allocation are not forming a loop (independent position) and total allocations are [math]m+n-1. [/math]This solution is termed as nin degenerated one.Thanks for suggestion and reading!