Let b be the price of 1 pound of beef, m be the price of 1 pound of mutton, and c be the price of 1 pound of crab.

From here, 3 statements can be established:

b+m=1040 --> Statement 1

2b+c=2120 --> Statement 2

2m+c=1708 --> Statement 3

Then, add Statements 2 and 3 together, to get: 2b+2m+2c=3828 --> Statement 4

For easier solving, divide Statement 4 by 2 to get: b+m+c=1914 --> Statement 5

You can now subtract Statement 1 from Statement 5, which will get you:

b+m+c-b-m=1914-1040, so

c=874

There, you have solved the price of a pound of crab. With this information, you can substitute c in Statement 2 to 874, so Statement 2 becomes:

2b+874=2120

2b=2120-874 =1246

b=623

Thus, the final answer: A pound of beef costs 623 cents.

let,price of one pound of beef is X,price of one pound of mutton is Y and price of one pound of crab is Z. now we can see that X + Y= 1040..............(1) 2X + Z = 2120.............(2) 2Y + Z = 1708..............(3)

   (2) - (3) we can get
   2X - 2Y = 412
  or, X - Y = 206...............(4)
  now from (1) + (4) we can get
  2X = 1246
  so, X = 623.
 so,The price of one pound of beef is 623 cents

Let us assume one pound of beef to cost $B$ cents, one pound of mutton to cost $M$ cents and one pound of crab to cost $C$ cents.

So, we have three conditions here. From them, we get three equations as follows:

$B + M = 1040$ ..... (1)

$2B + C = 2120$ ..... (2)

$2M + C = 1708$ ..... (3)

Solve the three equations for B,

we get

$B = 623$ cents

That's the answer!

LET

x = Beef

y = Mutton

z = Crab

X+Y = 1040

2X + Z = 2120

2Y + Z = 1708

then I minus the

II and III

2X - 2Y = 412

X - Y = 206

Then with the I I sum it

so

2X = 1246

X = 623

So the beef price of a pound of beef in cents is 623

Ofcourse they are using linear equations that contains 3 variables and it cant be solved until we dont have 3 equations. And already the butcher has given it to you. You represent cost of each pound of beef, mutton and crab with x, y and z respectively. You will get the following equations: x + y = 1040 2x + z = 2120 2y + z = 1708 In the first two equations try the simple elimination. By multiplying the Ist equation by 2 and then eliminating x you will get => "2y -z = -40" Then solve the resultant equation with the 3rd equation and you will get the value of y=417. Put this value of y in 1stequation and you will find the value of x= 623 which is for one pound of beef! And you are done!

tip: Never forget to take a pencil and paper with you. Who knows these math problem might pop up from somewhere! ;)

b+m =1040 (1)

2b+c =2120 (2) 2m+c =1708 (3)

2b-2m =412 (4) {subtract (3) from (2)}

2b+2m =2080 (5) {multiply (1) by 2}

add (4) & (5)

b = (412+2080)/4 =623

2b – 2m = 412
b – m = 206
b + m = 1040
2b = 1246
b = 623

X=Beef, Y=Mutton and Z=Crab
Put all known situation together
X+Y=1040
2X+Z=2120
2Y+Z=1708
Do it 1 by 1:
X+Y=1040 | 2|(Why? Because the amount of X in the 2nd situation is 2, so we need to time it by2)
2X+Z=2120| 1||(Why? Because the amount of X in the 1st situation is 1, so we need to time it by1) become=
2X+2Y=2080
2X+Z =2120
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ -(Minus)
We get 2Y-Z=-40
the 2nd situation with 3rd
2Y-Z=-40
2Y+Z=-1708
They no need to be timed because one of their amount has already same(1,2, variables etc, as long as one of them is equal, then no need to time)
2Y-Z=-40
2Y+Z=-1708
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ -(Minus)
-2Z=-1748
2Z=1748
Z=1748/2, 874

We just need to put the amount of Z in one of the equation that contains Z

2X+Z=2120,
2X+874=2120,
2X=2120-874,
2X=1246
x=623

B + M = 1040 ------------- (1)

2B + C = 2120 ...................(2)

2M + C = 1703 ...................(3)

(2) - (3) :

2B - 2M = 412

B - M = 206.................(4)

(1) + (4)

2B= 1246

B=623

1 pound of beef + 1 pound of mutton =1020 cents--------------------(1) 2 pounds of beef + 1 pound of crab =2120 cents--------------------(2) 2 pounds of mutton +1 pound of crab =1708 cents--------------------(3) solve 2 & 3 we get, 2 pounds(beef -mutton) = 412 cents-------(4) solve 1 & 4 we get, 2 pounds of beef =1246 cents hence 1 pound of beef = 623 cents

x+y=1040

2x+z=2120

2y+z=1708 *-1 =(-2y-z=-1708)

(2x+z=2120)+(-2y-z=-1708)=(2x-2y=412)

(x-y=206)+(x+y=1040)

2x=1246

x=623

B=1040-{1708-(2120-2B)/2} B=1040-(-206+B) B=1040+206-B 2B=1246 B=623

According to the problem, I determined 3 uknowns and 3 equations:

B = Beef M = Mutton C = Crab (Always 1 pound per each uknown.)

So:

B + M = 1040 -----> So, B = 1040 - M 2B + C = 2120 2M + C = 1708

Replacing B in the second equation:

2.(1040 - M) + C= 2120 2080 - 2M + C = 2120 - 2M + C = 40 C = 40 + 2M

Replacing C in the third equation:

2M + 40 + 2M = 1708 4M = 1668 M = 417

Finally, replacing M in the first equation:

B + 417 = 1040 B = 1040 - 417

B = 623 cents

B+M=1040............(1) 2B+C=2120...........(2) 2M+C=1708..........(3) Then eliminate equality (2) and (3) B-M=206...............(4) Then eliminate equality (1) and (4) B=623

x + y = 1040 ; 2x + k = 2120 ; 2y + k = 1708. So, 2(x+y) + 2 k = 3828 --> 2080 + 2k = 3828 --> k = 874 . Thus, 2x + 874 = 2120 --> 2x = 1246 ---> Answer : x = 1246/ 2 = 623

Let the cost of 1 pound of beef be x cents.

Therefore, cost of 2 pounds of beef = 2x cents

Let the cost of 1 pound of mutton be y cents.

Therefore, cost of 2 pounds of mutton = 2y cents

Let the cost of 1 pound crab be z cents.

According to first condition:

x + y = 1040 ----------- (i)

According to second condition:

2x + z = 2120 -----------(ii)

According to third condition:

2y + z = 1708 -----------(iii)

Subtracting (iii) from (ii):

2x - 2y = 412

2(x - y) = 412

x - y = 206 -----------(iv)

Adding (i) and (iv):

2x = 1246

x = 623

Hence the cost of 1 pound of beef is 623 cents.

Let us take price of 1 pound of beef as x, 1 pound of mutton as y and 1 pound of crab as z. Now, as given in the problem, we get the following equations ---

$x+y=1040$ ....(i) and $2y+z=1708$ ......(ii) and $2x+z=2120$ .....(iii)

On solving (i), we get $y=1040-x$ . Putting this value of y in (ii), we get,

$2y+z=1708 \implies 2(1040-x)+z=1708 \implies 2080-2x+z=1708 \implies z=(2x-372)$

Putting this value of z in (iii), we get,

$2x+z=2120 \implies 2x+(2x-372)=2120 \implies 4x-372=2120 \implies 4x=2492 \implies x=\boxed{623}$

let the cost of per pound beef be x , mutton be y and crab be z then, according to question, x+y=1040.... eq.(1) 2x+z=2120....eq(2) 2y+z=1708....eq(3) by eq(2) - z=2120-2x , substitute the value of z in eq(3) 2y+2120-2x=1708 we get , 2y-2x=-412.... eq(4) on adding eq(1)*2 and eq(4) (2y-2x=-412)+(2y+2x=2080) 4y=1668 y=417 now put value of y in eq(1) x+417=1040 x=1040-412 x=623 So, the cost of 1 pound of beef is 623 cents.

let pound of beef = x let pound of mutton = y let pound of crab = z 1 . x + y = 1040 2 . 2x+z= 2120 3. 2y+z= 1708 solving 2 and 3 we get x - y = 206 . . . .. (4) solving 1 and 4 we get x = 623