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Two weeks ago, you paid $55.30 for 10 apples, 8 canteloupes and 13 bunches of bananas. Last week, you paid $49.50 for 11 apples, 9 canteloupes and 10 bunches of bananas. This week you paid $57.50 for 11 apples, 7 canteloupes and 14 bunches of bananas. Question: How much does each type of fruit cost? [Write and solve a linear system.]



Answer :

msm555

Answer:

  • Cost of one apple ( x ): $0.90
  • Cost of one cantaloupe ( y ): $1.40
  • Cost of one bunch of bananas ( z ): $2.70

Step-by-step explanation:

Let's denote:

  • x as the cost of one apple,
  • y as the cost of one cantaloupe, and
  • z as the cost of one bunch of bananas.

We are given three sets of equations based on the purchases made over three weeks:

Two weeks ago:

[tex]\sf 10x + 8y + 13z = 55.30 [/tex]

Last week:

[tex]\sf 11x + 9y + 10z = 49.50 [/tex]

This week:

[tex]\sf 11x + 7y + 14z = 57.50 [/tex]

We can write this system of equations as follows:

[tex]\sf \begin{cases}10x + 8y + 13z = 55.30 \cdots (1)\\ 11x + 9y + 10z = 49.50 \cdots (2)\\ 11x + 7y + 14z = 57.50 \cdots (3) \\ \end{cases} [/tex]

To solve the system of equations, we can use the method of elimination. Let's start by manipulating the equations to eliminate one variable.

Subtract equation (1) from equation (2) to eliminate x :

[tex]\sf (11x + 9y + 10z) - (10x + 8y + 13z) = 49.50 - 55.30 [/tex]

Simplifying,

[tex]\sf x + y - 3z = -5.80 [/tex]

Subtract equation (1) from equation (3) to eliminate x :

[tex]\sf (11x + 7y + 14z) - (10x + 8y + 13z) = 57.50 - 55.30 [/tex]

Simplifying,

[tex]\sf x - y + z = 2.20 [/tex]

Now, we have a new system of equations:

[tex]\sf \begin{cases}x + y - 3z = -5.80 \\x - y + z = 2.20 \\\end{cases}[/tex]

Next, we can solve this simplified system by elimination or substitution.

Adding the two equations:

[tex]\sf (x + y - 3z) + (x - y + z) = -5.80 + 2.20 [/tex]

[tex]\sf 2x - 2z = -3.60 [/tex]

[tex]\sf x - z = -1.80 [/tex]

Now substitute x = -1.80 + z into one of the simplified equations. Let's use x - y + z = 2.20 :

[tex]\sf (-1.80 + z) - y + z = 2.20 [/tex]

[tex]\sf -1.80 + 2z - y = 2.20 [/tex]

[tex]\sf -y = 2.20 + 1.80 - 2z [/tex]

[tex]\sf -y = 4.00 - 2z [/tex]

[tex]\sf y = 2z - 4.00 [/tex]

Now substitute y = 2z - 4.00 and x = -1.80 + z back into one of the original equations to solve for z .

Let's use equation (1):

[tex]\sf 10(-1.80 + z) + 8(2z - 4.00) + 13z = 55.30 [/tex]

[tex]\sf -18 + 10z + 16z - 32 + 13z = 55.30 [/tex]

[tex]\sf 39z - 50 = 55.30 [/tex]

[tex]\sf 39z = 105.30 [/tex]

[tex]\sf z = \dfrac{105.30}{39} [/tex]

[tex]\sf z \approx 2.70 [/tex]

Now substitute z = 2.70 back into y = 2z - 4.00 and x = -1.80 + z to find y and x :

[tex]\sf y = 2(2.70) - 4.00 = 5.40 - 4.00 = 1.40 [/tex]

[tex]\sf x = -1.80 + 2.70 = 0.90 [/tex]

Therefore, the solution to the system is:

[tex]\sf x = 0.90, \quad y = 1.40, \quad z = 2.70 [/tex]

Hence, the cost of each type of fruit is:

  • Cost of one apple ( x ): $0.90
  • Cost of one cantaloupe ( y ): $1.40
  • Cost of one bunch of bananas ( z ): $2.70

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