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Sagot :
Let's look at an example.
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Problem:
At a high school play, tickets are sold at two different prices: student prices and parent prices. The student price is $2 while the parent prices is $3. On one night, 145 tickets were purchased and the total revenue was $335.
How many student tickets were purchased? How many parent tickets were purchased?
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Solution:
Let x be the number of student tickets sold and y be the number of parent tickets sold.
We know that 145 people attended so x+y = 145 is the first equation we can set up. This equation is in standard form since standard form is Ax+By = C. Here we have A = 1, B = 1, C = 145.
The second equation we can form is 2x+3y = 335. This is also in standard form. This time we have A = 2, B = 3, C = 335.
The 2x is the amount of money made from just the students only. The 3y is from the parents only. The expression 2x+3y is the total from both groups. This total is set equal to 335 as this is the amount of money pulled in.
After forming both equations, we can write this system of equations
[tex]\begin{cases}x+y = 145\\2x+3y = 335\end{cases}[/tex]
To solve this system, we could isolate y in the first equation to get y = -x+145. Then plug this into the second equation to solve for x
2x+3y = 335
2x+3(-x+145) = 335
2x-3x+435 = 335
-x = 335-435
-x = -100
x = 100 .... number of student tickets bought
Use this to find y
y = -x+145
y = -100+145
y = 45 .... number of parent tickets bought
We can see that x+y = 100+45 = 145
And also 2x+3y = 2*100+3*45 = 200+135 = 335
These two facts help confirm we have the correct x,y values.
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