If x+y=30 and x-y=2, Find x and y.

i know the answer but lets see how many ppl can answer this.

(Level: Novice)

Hint: Solve using Simultaneous Equations.

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x + Y = 30 ... (Equation A) and

x - y = 2 ... (Equation B)

We can solve this question by several methods.

**Method 1**

Adding these two equations, that is equations A and B, we get:

(x + y) + (x - y) = 30 + 2

(please note that for addition of equations, we add left hand side and right hand side of equalities separately.

Therefore: 2x = 32

Therefore: x = 32/2 = 16

Substituting this value of x in first equation we get:

16 + y = 30

Therefore: y = 30 - 16 = 14

Thus the answer is: x =16 and y = 14

Method 2

Transferring y in equation B from left hand side to right hand side we get:

x = y + 2

Substituting this value of x in equation A we get:

(y + 2) + y = 30

Therefore: 2y + 2 = 30

Therefore: 2y = 30 - 2 = 28

Therefore: y = 28/2 = 14

Substituting this value of y in equation A we get

x + 14 = 30

Therefore: x = 30 - 14 = 16

No algebra!

First equation means like:Two people x and y get a sum of 30 dollars.

The second of the equations means x, one of the two persons, gets 2 dollars more than the person y.

x and y are names of two persons chosen from your equations. Dollar is only for making a meaningful practical example of your equatios.

Solution:Keep the 2 dollars, that x should get more than y, separate. The rest of the 30-2=28 dollars, then, have to be shared equally between x and y : So, 28/2 = **14** that is the value for** y.**

x should be 2 more than y . So give that 2dollars separated to x which makes his share : x=14+2=**16. **This is a commonsense or a method of logic.

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Algebraically you can do by elimination of y from the two given simultaneous equations:

x+y=30 (1)

x-y=2 (2)

There are 2 equalities (or equations). By the postulate that equals can be added to equals and the equality remains we perform the following:

Adding the given two equations we get 2x=30+2=32.

Therefore,x=32/2=16.

Subtracting the equation (2) from (1), we get x eliminated and only yor ys remains on the left:

2y=30-2=28

y=28/2=14.

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Also x **+y =30** and x ** - y = 2** means that on a number line x is the middle of the stretch of 2 to 30 and so, x = 16, is the middle. The equidistance on either side of (x=16) is 14 , for this equal distance from x=16, we get 2 on left and 30 on right. By this interpretation **x=16** and **y=14** are the solution of these two simultaneous equations. Here is a rough picture for the imagination only but please do not try measuring it.

line: |-----y-----------| |---------y--------| line

(x-y)=2|----------------(x=16)-----------------30=(x+y)

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You can use graph to solve the two equations.

x+y=30 is a many valued equation,representinga straight linecutting x axis at 30 and y axis at 30 units from the origin.

x-y=2 also represents an equation with many solutions, representing a straight line cutting x axis at x=2 and y axis at y=-2.

Both straight lines you can plot and they are perpendicular to each other intersecting at **x=16** and **y=14** which is the common (or simutaneous unique solution for both equations)

Please plot the following and see where the two graphs cut.

x+y=30 or y= x-30

x=00 02 04 06 08 10 12 14 **16** 18 20 22 24 26 28 30

y=30 28 26 24 22 20 18 16 **14** 12 10 08 06 04 02 00

x-y=2 or y=x-22:

x=+00 02 04 06 08 10 12 14 **16 **18 20 22 24 26 28 30

y=-02 00 02 04 06 08 10 12 **14** 16 18 20 22 24 26 28

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By trial and error way we can come around the solution.

Choose any two numbers such that their sum is 30. Say 10 and 20.The difference gap is 10. Then try reducing their gap by inreasing the smaller. You increase the smaller ,10 by one , i.e 10+1 =11 and decrease 20 by one , i.e to 19. Now the gap is reduced from 10 to 9. Inrease by one by one the lower number and decrease the greater number one by one till their difference of two resulting number become 2. The number finally is 14 and 16. 16 is ,therefore, x and cosequently, 14 is y

Hope this helps.

x+y=30 - equation 1

x-y=2 - equation 2

Next, subtract equation 1 to equation 2

x-x+y-(-y)=30-2

2y=28

y=14

Next, substitute y=14 into equation 2

x-(14)=2

x-14=2

x=14+2

=16

*x=16, y=14

Hope it answers the questions

x + Y = 30 (Equation A) and

x - y = 2 (Equation B)

Honestly, you could solve by using logic. Looking at Equation A, you know that x + y = 30 and Equation B tells you x is 2 more than y. So what two number give you a sum of 30, when one of them is 2 more than the other?

It can't be 15 + 15 since their difference is 0, however, it can be 16 + 14 since 16 is 2 more than 14.

So, x - y = 2 is the equivalent as 16 - 14 = 2

And they add up to give you 30. 16 + 14 = 30

x = 16

y= 14

Hope that wasn't confusing ;)

Uh...everyone can answer this...?

30-2=2y, y=14, x=16

a three second thought.

x-y=2 To let x by itself you have to add y on both side

x-y+y=2+y

x=2+y You pug 2+y in the x in this equation x+y=30

2+y+y=30

2+2y=30 You subtract 2 on both side

2+2y-2=30-2

2y=28 You divide both side by 2 to let y by itself

2y/2=28/2

**y=14**

To find x you just have to plug 14 for y

x-14=2 add 14 on both side

x-14+14=2+14

**x=16**

x+y=30 and x-y=2

x = y + 2

y + y + 2 = 30

2y + 2 = 30

2y = 28

y = 14

x - 14 = 2

x = 16

x+y=30 and x-y=2

x-y=2 add y

x = 2+y

plug it in as x in the other problem

2 + y + y =30

combine like terms

2 + 2y = 30

subtract 2

2y = 28

divide by 2

y=14

now plug in the y into the other problem to find x

x- 14 = 2

add 14 to both sides

x = 16

so **y = 14** and **x = 16**

IF, X+Y=30

X=30-Y......... EQN(i)

SIMILARLY,

IF,X-Y=2

X=2+Y...........EQN(ii)

FROM EQN(i) & EQN(ii)

30-Y=2+Y

30-2=Y+Y

2Y=28

Y=14

NOW,

PUTTING THE VALUE OF Y IN EQN(i),

X=30-Y

X=30-14

X=16

HENCE, X=16,

Y=14

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