Linear systems?

Question

Three footballs and one soccer ball cost $155. Two footballs and three soccer balls cost $220. Determine the cost of one football and the cost of one soccer ball.

Answer 1

Let us take the football as X and soccer ball as Y. So according to the conditions given above

3X + Y = 155 ........(lets call this as eq 1)
2X + 3Y = 220 ........(lets call this as eq 2)

Now v hv to get rid to either the X or the Y to get an ans. So v hv to hv the same co-efficient for X or Y. Let us take Y. To hv the same coefficient v will hv to multiply the eq 1 by 3, and then subtract it from eq 2

9X + 3Y = 465 ..... (multiplying it by 3)
2X + 3Y = 220 ......(the original eq 2)

Now subtract the two eq to get a value for X. Notice that Y gets cancelled.

7X = 245

X = 245 / 7

X= 35
Since v now hv the value of X, just put back the value in either eq 1 or in eq 2 to get Y. Lets put in eq 1

3X + Y = 155

3*35 + Y = 155 .........( X is 35)

105 + Y = 155

Y = 155 - 105

Y=50

Hence X = 35, Y = 50

Hence each football = $35, soccerball = $50
(U can cross check if the values are right by just putting the value for X and Y in either of the eq, if the value is same as on the right u r correct)

Answer 2

F = 35
S = 50

3f + 1s = 155
3f + s = 155
s = 155 -3f ---> price of 1 soccer ball


2f + 3f = 220
2f + 3(155-3f) = 220
2f + 465 - 9f = 220
465 - 220 = 9f - 2f
245 = 7f
f= 245 / 7
f = 35
each football costs $35

so now...
f = 35 and
s = 155 - 3f
s = 155 - 3 (35)
s = 155 - 105
s = 50
each soccerball is $50

Answer 3

Let F = the cost of a football
Let S = the cost of a soccer ball

3F + 1S = 155
2F + 3S = 220

I leave it to you to solve the system of equations.