Chantrea bought 8 apples and 5 pears for $5.72. At the same store, Dara bought 6 apples and 2... Question: Chantrea bought {eq}8 {/eq} apples and {eq}5 {/eq} pears for {eq}$5.72 {/eq}. At the same store, Dara bought {eq}6 {/eq} apples and {eq}2 {/eq} pears for {eq}$3.38 {/eq}. What is the cost of {eq}1 {/eq} apple at this store? Elimination Method: (i) The elimination method is used to solve a system of two equations in two variables. (ii) In this method, we add or subtract the given equations and eliminate one variable. (iii) We solve the resultant equation in one variable using the algebraic operations. Answer and Explanation: Let us assume that the costs of each apple and each pear are {eq}x {/eq} dollars and {eq}y {/eq} dollars respectively. The problem says, "Chantrea bought {eq}8 {/eq} apples and {eq}5 {/eq} pears for {eq}$5.72 {/eq}".

So we get the equation:

$$8x+5y= 5.72 \,\,\,\,\,\,\,\rightarrow (1)$$

The problem also says, "Dara bought {eq}6 {/eq} apples and {eq}2 {/eq} pears for {eq}$3.38 {/eq}". So we get the equation: $$6x+2y = 3.38\,\,\,\,\,\,\,\rightarrow (2)$$ Multiply equation (1) both sides by {eq}-2 {/eq}. Then we get: $$-16x-10y=-11.44 \,\,\,\,\,\,\,\rightarrow (3)$$ Multiply equation (2) both sides by {eq}5 {/eq}: $$30x+10y =16.9\,\,\,\,\,\,\,\rightarrow (4)$$ Adding (3) and (4): $$(-16x-10y) + (30x+10y) = -11.44 +16.9 \\[0.4cm] \text{Combining the like terms}, \\[0.4cm] (-16x+30x)+(-10y+10y) = 5.46 \\[0.4cm] 14x = 5.46 \\[0.4cm] \text{Dividing both sides by 14}, \\[0.4cm] x = 0.39$$ Therefore, the cost of each apple = {eq}\boxed{\mathbf{\$ 0.39}} {/eq}.

Elimination Method in Algebra: Definition & Examples

from High School Algebra II: Help and Review

Chapter 7 / Lesson 9
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