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I Need To Know How To Do A/x - B/y = 1 Solve For X.

I need to know how to do: a/x - b/y = 1 solve for x.?

To solve for x just move stuff around until x is by itself on one side.

Start with adding b/y to both sides of the equation. If two things equal one another and you add the same amount to both then they will still equal each other.

a/x - b/y + b/y = 1 + b/y

or a/x = 1 + b/y

I prefer x to be the numerator so lets take the inverst of both sides of the equation. Again, doing the same thing to both sides keeps the equals sign intact.

1/(a/x) = 1/(1 + b/y)

or x/a = 1/(1 + b/y)

now multiply both sides by a.

x=a/(1/(1 + b/y)

4x+2y=98 i need to solve for x and y?

there are two way s to approach a problem like this...

1) you need another equation of x and y ... if you have that then you solve them simultaneously and the answer is A point where the two lines intersect.

2) you play the guessing game. you guess a X value and then calculate an y (or guess a Y and calulate an X) usually it is called a T chart something like this


X ( guess)............................... Y (calculate)
0................ then 2y=98, y = 98/2, y= 49
-1................then -4 +2y =98, 2y= 98+4, 2y=102, y=102/2, y=51
1.................the 4 +2y =98, 2y=98-4, 2y = 94, y=94/2, y= 47


so possible answers are (0, 49),(-1,51),(1, 47) the list of possibilities is endless

How do you solve 6 + x = 1?

In an equation both sides of the equation equal each other. You can multiply, divide, add or subtract sides in an equAtion as long as you do the exact same thing on another side and as long as you do that then the equation remains true. Take the 6=6 equation, if you subtract 1 on both sides it becomes 5=5 and if you then multiply 2 on both sides then it becomes 10=10 which means the equation is still correct. If you just multiply by 2 on the left side then you get 10=5 which doesn’t make any sense. So you can really do whatever you want to an equation as long as you ensure that Both sides remain equal to one another. In knowing that you can take:6 + X = 1 and if you want to know what X is then you need to have X on one side by itself so that the number can be on the other side. So if we take away or subtract 6 from 6 + X, it looks like this : 6 + X - 6 , now 6–6 = 0, so now you have X on one side. Now remember, what we subtract on one side of this equation we must subtract on the other side, so: 1 - 6 = - 5. Let me show you again with both sides of the eqution:6 + X -6 = 1 -6so now we have X on one side and -5 on the other side.Which means the equation is still true andX = - 5. …..now you can test this by going placing -5 back into the original equation.6 + -5 = 6 - 5 = 1 (remember ( + - ) = (-) )

How do you solve 7 - x = 3?

7 - x = 3add x on both sides7 - x + x = 3 + xso7 = 3 + xsubtract - 3 on both sides7 - 3 = 3 - 3 + xso4 = xorx = 4_________________________________________________________You can follow me here:Facebook: Khuram Waqas Ur RehmanTwitter: Khuram Waqas (@khuramwaqas) | TwitterInstagram: Khuram Waqas Ur Rehman (@kwrehman) • Instagram photos and videosGoogle plus: Khuram Waqas Ur Rehman - Google+Ask.fm: Khuram Waqas Ur Rehman (@Kwrehman). Ask me anything on ASKfmLinkedin: http://www.linkedin.com/in/khuramwaqasVisit my blogs here :Thoughts and RealityDaily News UpdateWatch this video and subscribe my channel :negative impact of technology on kids

How do you solve the inequality 1/x < 4 ?

(-infinity, 0) and (1/4, infinity) is correct.

When solving an inequality with x as a denominator, you know that x cannot be zero. However, when x is negative, then you have to switch the sign of the inequality when you multiply both sides by x. You set up to cases, as shown here:


CASE I - x is a positive number

1/x < 4
1 < 4x
1/4 < x


CASE II - x is a negative number

1/x < 4, x is not 0.

Looking at the equation itself, any negative number you plug in for x will turn the entire quantity into a negative number, which by definition is less than 4. Therefore any negative number works.


So your final intervals are (-infinity, 0) and (1/4, infinity)

How do i solve x - y = -1 and x + y = 5?

For the best answers, search on this site https://shorturl.im/axahy

@@@ (1/x) - (1/y) = (1/3) ... (1/x²) + (1/y²) = (5/9) (1/x) = A ... (1/y) = B A - B = (1/3) A² + B² = (5/9) ( A - B )² = (1/9) A² - 2AB + B² = (1/9) (5/9) -2AB = (1/9) (4/9) = 2AB ( A + B )² = A² + 2AB + B² ( A + B )² = (5/9) + (4/9) =1 (1) when A + B = 1 A + B = 1 A - B = (1/3) 2A = (4/3) ... A = (2/3) ... B = (1/3) ( x , y ) = ( 3/2 , 3 ) valid (2) when A + B = -1 A + B = -1 A - B = (1/3) 2A = (-2/3) ... A = (-1/3) ... B = (-2/3) ( x , y ) = ( -3 , -3/2 ) valid @@@ addition (1/x) + (-1/y) = (1/3) ... (1/x²) + (1/y²) = (5/9) α=(1/x)...β=(-1/y) α+β=(1/3)...α²+β²=(5/9) αβ=[(α+β)²-(α²+β²)]/2    =[(1/9)-(5/9)]/2=(-2/9) t²-(1/3)t-(2/9)=0 [ t-(2/3) ][t+(1/3)]=0 (1) α=(2/3) , β=(-1/3)   (x,y)=( 3/2 , 3 ) (2) α=(-1/3) , β=(2/3)   (x,y)=( -3 , -3/2 ) @@@

How do we solve [math]x + 2 = x^2[/math]?

How do we solve [math]x+2 = x^2[/math]?By just looking at the equation, you may guess it that one of the answers to that question is [math]x = 2[/math]. But can you also figure out that [math]x=-1[/math] is also the answer?When we deal with a quadratic equation like this, it is usually safe to assume that there are two solutions to the problem. Not all of the quadratic equations have two solutions. Some of them have only one solution (consider [math]x^2=0[/math]), some of them have no real solutions (consider [math]x^2=-1[/math]). But lots of them have two solutions, just like this:[math]x^2=9[/math]The solution to the quadratic equation above is x=3 and x=-3 (never forget that -3 to the power of 2 is also 9).So, back to the problem in the question.[math]x+2 = x^2[/math]How do we solve that? One of the simplest ways is to move all of the variables and the constant to one side, so we get zero on the other side.[math]x^2 - x - 2 = 0[/math]Can you factorize the quadratic expression on the left hand side? If you don’t know how to factorize, you may want to google it.[math](x+1)(x-2)=0[/math]When you see this form, you know that either [math](x+1)=0[/math] or [math](x-2)=0[/math] because for AB=0, it must be either A=0 or B=0 (or both).Therefore, [math]x=-1[/math] OR [math]x=2[/math].

How do you solve x + 3 = 7?

The equation is x + 3 = 7Okay, since you guys may need to understand this through an exampleImagine that you and your friend have a few bottles of coke each. Both of you know that you have 7 bottles, but you don't know how many does your friend has. He tells you that he needs 3 more bottles of coke to match up to you. Now you buy 3 bottles for him and give it to him. You know that now you both have equal number of bottles. Therefore as you have 7 bottles even your friend has seven bottles. Now you wonder how many did he manage to collect on his own. You only know that you had 7 bottles and he added 3 bottles to his collection to make it 7.So if we assume that he had x number of bottles and we don't know that what is the value of x ( is it 2, 3 or 8). So you know that you had 7 and your friend also had 7 but out of that, you give him 3 of them.Therefore you write it asHis collection was less than you collection but you gave him more and made it even.So now he has 3 more than x and you have 7. Also you know that when you add 3 more to x, it becomes 7.Therefore -X+3=7Now you take out 3 from his collection. So, his total number of bottles also get reduced by 3Therefore -X+3–3 =7–3This becomes x = 4Therefore, your friend was only able to collect 4 bottles of coke where as you collected 7.So now through a fun way(I hope) you know the answer to your question

How do I solve a 4*4 Rubik's Cube?

There are many methods to solve a 4x4 Rubik’s cube. The method I follow is inspired from solving the 3x3.The main difference between a 3x3 and a 4x4 is that the centers of the 4x4 are not fixed unlike a 3x3. Therefore, I suggest you to build the center(the middle 2x2) of the 4x4.While building the centers make sure the colors are oriented in the right manner as it would be in a 3x3. Else, you might find some difficulties.Now solve the two edge pieces, bring all identical edge pieces together.Now choose one layer, I generally go with the white, and build the cross with the aligned edge piecesNow try to complete that layer(ex: white) by filling in the corner pieces. While doing so, you will automatically solve the three layers and will be left with just one layer.If you do not end up with a parity, you can now use same methods used in a 3x3 to complete the cube. If you do end up in a parity, there are steps to break the parity as well.To understand better I suggest this youtube video:

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