Matrix word problems

Question

Brenda's school is selling tickets to a spring musical. On the first day of ticket sales the school sold 3 senior citizen tickets and 9 child tickets for a total of $75. The school took in $67 on the second day by selling 8 senior citizen tickets and 5 child tickets. What is the price each of one senior citizen ticket and one child ticket?

Answer 1

let x be the price of a senior ticket and let y be the price of a child ticket

on day 1, we have

3x + 9y = 75

on day 2, we have

8x + 5y = 67

I want to make the x in the second equation 0. I see that there are 8 xs in that equation, and 3 in the first equation. So, I multiply the first equation by (8/3) to make it have 8 xs also. Then I subtract equation 2 by that.

equation 1 times (8/3):

(8/3)(3x + 9y) = (8/3)75

simplifying:

8x + 24y = 200

equation2 - (8/3)equation1:

(8x + 5y) - (8x + 24y) = 67 - 200

simplifying:

-19y = -133

y = 7

now we plug y = 7 into the first equation. this is called "back substitution".

3x + 9*7 = 75

3x + 63 = 75

3x = 12

x = 4

so, senior tickets cost $4 and child tickets cost $7. we can check this makes sense: senior tickets are cheaper, and plugging everything into equation 1: 3*4 + 9*7 = 75

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now i will repeat the same solution as a matrix equation, just so you see how the notation works.

we have the equation:

[3 8] [x] .. [75]

[8 5] [y] = [67]

(you can ignore the dots. it won't allow me to put multiple spaces)

which we can solve this by forming the so called augmented matrix:

[3 8 | 75]

[8 5 | 67]

and putting this in "row echelon form". This takes only one step here: subtract (8/3) row 1 from row 2.

[3 8 | 75]

[0 -19 | -133]

I solve the equations from the bottom rows upwards using backsubstitution. the bottom row gives 0x -19y = -133. That gave y=7. Then I plug y=7 into the top row, solving for x. That gave x=4.