Answer:
(i) One apple = $0.40
(ii) One banana = $0.30
Step-by-step explanation:
Given information:
- Cost of 4 apples and 3 bananas = £2.50
- Cost of 3 apples and 4 bananas = £2.40
Define the variables:
- Let x = The cost of one apple.
- Let y = The cost of one banana.
Create two equations with the given information and defined variables:
[tex]\textsf{Equation 1}: \quad 4x+3y=2.50[/tex]
[tex]\textsf{Equation 2}: \quad 3x+4y=2.40[/tex]
Multiply the first equation by 4:
[tex]\implies 4 \cdot 4x+4 \cdot 3y=4 \cdot 2.50[/tex]
[tex]\implies 16x+12y=10.00[/tex]
Multiply the second equation by 3:
[tex]\implies 3 \cdot 3x + 3 \cdot 4y = 3 \cdot 2.40[/tex]
[tex]\implies 9x + 12y = 7.20[/tex]
Subtract the equations to eliminate the term in y:
[tex]\begin{array}{crcccc}& 16x & + & 12y & = & 10.00\\-& (9x & + & 12y & = & 7.20)\\\cline{2-6} & 7x & & &=&2.80\\\end{array}[/tex]
Solve for x:
[tex]\implies 7x=2.80[/tex]
[tex]\implies \dfrac{7x}{7}=\dfrac{2.80}{7}[/tex]
[tex]\implies x=0.40[/tex]
Therefore, the cost of one apple is $0.40.
Substitute the found value of x into one of the equations and solve for y:
[tex]\implies 4x+3y=2.50[/tex]
[tex]\implies 4(0.40)+3y=2.50[/tex]
[tex]\implies 1.60+3y=2.50[/tex]
[tex]\implies 1.60+3y-1.60=2.50-1.60[/tex]
[tex]\implies 3y=0.90[/tex]
[tex]\implies \dfrac{3y}{3}=\dfrac{0.90}{3}[/tex]
[tex]\implies y=0.30[/tex]
Therefore, the cost of one banana is $0.30.
