Lena runs 2/3 mile each day. She wants to know how many days she had to run before she had run a whole number of miles?

Tickets to a musical cost $30 for adults and $12 for children. At one particular performance 960 people attended and $19 080 was

postnew [5]

In order to solve this, you have to set up a systems of linear equations.

Let's say that children = c and adults = a

30a + 12c = 19,080
a + c = 960

I'm going to show you how to solve this system of linear equations by substitution, the easiest way to solve in my opinion.

a + c = 960
- c - c
---------------------- ⇒ Step 1: Solve for either a or c in either equation.
a = 960 - c

20(960 - c)+ 12c = 19,080
19,200 - 20c + 12c = 19,080
19,200 - 8c = 19,080
- 19,200 - 19,200
---------------------------------- ⇒ Step 2: Substitute in the value you got for a or c
8c = -120 into the opposite equation.
------ ---------
8 8

c = -15

30a + 12(-15) = 19,080
30a - 180 = 19,080
+ 180 + 180
-------------------------------
30a = 19,260
------- -----------
30 30

a = 642
__________________________________________________________

I just realized that there can't be a negative amount of children, so I'm sorry if these results are all wrong.

7 0

2 years ago

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At the county fair, the operator of a game guesses a contestant’s weight. For each pound the operator’s guess differs from the c

BartSMP [9]

Answer: $| x - \frac{15}{3} | = 120$ or $\frac{1}{3}| 3 x - 15|= 120$ or $| 3x - 15| = 360$ or $| x - 5 | = 120$