We went to the market to buy some fruit. We decided to buy some apples, strawberries, and oranges. If we buy 2 apples, 3 boxes of strawberry, and 4 oranges, the fruit would cost $15.30. If we buy 1 box of strawberry, 4 apples, and 2 oranges, the fruit would cost $10.90. If we buy 1 orange, 5 apples and 2 boxes of strawberry, the fruit would cost $13.70. Use the Gaussian Elimination Method to solve for the price of each fruit.

Step-by-step explanation:

Let the amount of apples bought be x

Amount of strawberries bought be y and;

amount of oranges bought be z

If we buy 2 apples, 3 boxes of strawberry, and 4 oranges, the fruit would cost $15.30, then this will be expressed as;

2x+3y+4z = 15.30 1

If we buy 1 box of strawberry, 4 apples, and 2 oranges, the fruit would cost $10.90, this will be expressed as:

4x+y+2z = 10.90 2

If we buy 1 orange, 5 apples and 2 boxes of strawberry, the fruit would cost $13.70, this is expressed as:

5x+2y+x = 13.70 3

Solving the three equations simultaneously:

2x+3y+4z = 15.30 1

4x+y+2z = 10.90 2

5x+2y+z = 13.70 3

Reduce the number of equations

multiply equation 1 by 2 and subtract from equation 2

equation 1 * 2 will give:

4x+6y+8z = 30.6 4

eqn (4)-eqn(2)

6y-y + 8z-2z = 30.6-10.90

5y+6z = 19.7 5

Also multiply equation 2 by 5 and eqn 5 by 4 and subtract from each other.

eqn(2)* 5 will give:

20x+5y+10z = 54.5 *

eqn(3) * 4 will give

20x+8y+4z = 54.8**

Subtsrct ** from *

8y-3y+(4z-10z)= 54.8-54.5

5y-6z = 0.3 6

Equate 5 and 6 and solve:

5y+6z = 19.7 5

5y-6z = 0.3 6

subtract:

6z+6z = 19.7+0.3

12z = 20

z = 20/12

z = 1.67

Substitute z = 1.67 into eqn 6

5y-6(1.67) = 0.3

5y - 10 = 0.3

5y = 10.3

y = 10.3/5

y = 2.06

Substitute z = 1.67 and y = 2.06 into equation 1

From 1:

2x+3y+4z = 15.30

2x+3(2.06)+4(1.67) = 15.30

2x+6.18+6.68 = 15.30

2x+12.86 = 15.30

2x = 15.30-12.86

2x = 2.44

x = 2.44/2

x = 1.22

Therefore 1 apple costs $1.22, 1 strawberry costs $2.06 and 1 orange costs $1.67