A local high school held a two-day fundraiser selling lunches. The equations shown represent the amount of money the school coll
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Answer:
a = $13.5
Step-by-step explanation:
Let a = adult tickets
Let c = children tickets
Translating the word problem into an algebraic equation;
<u>For the Martinez family;</u>
2a + 3c = $60
<u>For the Wright family;</u>
3a + 5c = $95.5
Thus, the simultaneous equations are;
..........equation 1
.........equation 2
We would use substitution method to solve;
From equation 2, we make a the subject of formula;
3a = 95.5 - 5c
a = (95.5 - 5c)/3
<em>Substituting the value of "a" into equation 1, we have;</em>
2[(95.5-5c)/3] + 3c = 60
Multiplying all through by 3;
2(95.5 - 5c) + 9c = 180
191 - 10c + 9c = 180
191 - c = 180
c = 191-180
c = $11
To find the value of a;
2a +3c = 60
<em>Substituting the value of "c" into the equation, we have;</em>
a = $13.5
<em>Therefore, the cost of an adult movie ticket is $13.5. </em>