SOLUTION: A school offers a total of 9 different Latin and German courses. If the number of German courses offered at the school Is 3 less than twice the number of Latin classes, h / many La
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Question 599145: A school offers a total of 9 different Latin and German courses. If the number of German courses offered at the school Is 3 less than twice the number of Latin classes, h / many Latin classes are offered at the school? Answer by math-vortex(648) (Show Source):
You can put this solution on YOUR website! Hi, there---
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To solve this problem, we write and solve a system of equations describing the given relationships.
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[I] Define your variables
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Let L be the number of Latin courses, and let G be the number of German courses.
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[II] Write a system of equations
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The total number of classes (Latin plus German) is 9. In algebra, we write
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The phrase, "the number of German courses...is 3 less than twice the number of Latin courses" can be written as
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[III] Solve the system of equations
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Re-write the first equation in terms of G.
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Substitute 9-L for G in the second equation.
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Simplify and solve for L. Add 3 to both sides of the equation.
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Add L to both sides of the equation.
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Divide both sides by 3 to isolate the variable L.
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The equation L=4 means that there are 4 Latin courses. Thus there are 9-4=5 German courses (since 4+5=9.)
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[IV] Check your work.
Is the number of German classes 3 less than twice the number of Latin courses? Twice the number of Latin courses is 4*2=8 courses. Five is 3 less than eight.
Check!
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There are four Latin courses and five German courses at the school.
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That's it! Feel free to email me via gmail if you have questions about the solution.
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Ms.Figgy
math.in.the.vortex@gmail.com
