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Let a = number of adult tickets. Let s = number of student tickets. She sold a total of 50 tickets. Can you write an equation that shows the total number of adult and student tickets is 50 using variables a and s?
A+s=50 is it like this
Great. We have one equation: a + s = 50 Now we need a second equation. We will get it from the prices of the tickets and the total amount of money collected.
A5+S3=180 then how do i solve those two equation
Each adult ticket costs $5, and a tickets were sold. How much money do a tickets at $5 each cost? Then we do the same for student tickets. s number of student tickets were sold at $3 each. What is the cost of s tickets at $3 each?
i am kind of confused now
Ok, you got it already. We have this system of equations: a + s = 50 5a + 3s = 180 Is there any method of solving a system of equations you prefer, substitution or addition?
Let's use substitution. It's a very common method of solving a system of equations.
but now how do i solve these two equation
In substitution, first you solve one equation for one variable. Then you substitute that into the other equation.
Let's solve the first equation for a. a + s = 50 If we subtract s from both sides, we'll get a by itself. That's what we want.
a + s = 50 -s -s ---------------- a = -s + 50
Now that we have a = -s + 50, we substitute that into the second equation. Where we see a in the second equation, we replace it with -s + 50. 5a + 3s = 180 5(-s + 50) + 3s = 180 <------ This is the substitution step.
then how do i solve 5[-s+50]+3s=180
Now we have an equation with only one variable, s, so we can solve for s. 5(-s + 50) + 3s = 180 Distribute the 5: -5s + 250 + 3s = 180 Combine -5s and 3s: -2s + 250 = 180 Subtract 250 from both sides: -2s = -70 Divide both sides by -2: s = 35. Now that we have s = 35, we substitute this value into the first original equation and solve for a: a + s = 50 a + 35 = 50 Subtract 35 from both sides: a = 15 We get a = 15 and s = 35. Since this was a word problem, and we were asked a number of tickets for the adults and students, we answer with a statement: She collected 15 adult tickets and 15 student tickets. (We should not hand in an answer of a = 15 and s = 35 because the variables a and s were used by us to solve the problem, but they are not stated in the problem.)
how did u distrute the 5 from 5[-s+50]+3s=180
and why did u combined -5s and 3s
The distributive property is: a(b + c) = ab + ac We had 5(-s + 50), so following the pattern of the distributive property above we get: 5(-s + 50) = (5) * (-s) + 5 * 50 = -5s + 250 Since that was part of an equation, the rest of the equation was still there: 5(-s + 50) + 3s = 180 -5s + 250 + 3s = 180
When you are solving an equation with one variable, you need to combine all terms that have that variable. In our case, we had -5s + 250 + 3s = 180 We need all terms with s combined on the left side, and all numbers on the right side, so that in the end we have only s on the left side equaling a number. We combine terms with s: -5s + 250 + 3s = 180 -5s + 3s + 250 = 180 -2s + 250 = 180 Then we subtract the number 250 to the right side, so we have only the terms with s on the left side and the numbers on the right side. -2s = -70 Now we do have s on the left side, but it is still being multiplied by -2. We want s by itself, so we divide both sides by -2 s = 35
i have one more question left why did u distribute?
And thank u so much u really help me a lot.
Since you can't combine unlike terms -s and 50, you need to distribute to get rid of the parentheses.