Consider the algebraic expression (a+b)^10
1 answer:
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Cost of 1 apple is $ 1.5 and cost of 1 pear is $ 1.25
<h3><u>Solution:</u></h3>Let "a" be the cost of 1 apple
Let "p" be the cost of 1 pear
Given that,
<em><u>One week Beth bought 3 apples and 8 pears for 14.50</u></em>
So we can frame a equation as:
3 apples x cost of 1 apple + 8 pears x cost of 1 pear = 14.50
3a + 8p = 14.50 ----- eqn 1
<em><u>The next week she bought 6 apples and 4 pears and paid 14$</u></em>
So we can frame a equation as:
6 apples x cost of 1 apple + 4 pears x cost of 1 pear = 14
6a + 4p = 14 ---- eqn 2
<em><u>Let us solve eqn 1 and eqn 2 to find values of "a" and "p"</u></em>
Multiply eqn 2 by 2
12a + 8p = 28 ---- eqn 3
Subtract eqn 1 from eqn 3
12a + 8p = 28
3a + 8p = 14.50
(-)----------------
9a = 13.5
<h3>a = 1.5</h3>From eqn 1,
3a + 8p = 14.50
3(1.5) + 8p = 14.50
4.5 + 8p = 14.50
8p = 10
<h3>p = 1.25</h3><em><u>Thus we have:</u></em>
Cost of 1 apple is $ 1.5 and cost of 1 pear is $ 1.25