# How construct simultaneous equations from worded problems?

## Can someone please explain to me how to do question 17 and 14? Answer this and I will love you forever!

Feb 2, 2017

Problem 17 set up:

You are given that there are 3 types of tea.
Let x = the mass (in kilograms) of the tea for $10 per kilogram Let y = the mass (in kilograms) of the tea for$11 per kilograms
Let z = the mass (in kilograms) of the tea for $12 per kilograms The phrase "so as to obtain 100 kilograms" allows us to write equation [1]: $x + y + z = 100$kilograms [1] The phrase "100 kilograms worth$11.20 per kilograms" means that the blend will cost:

(100" kilograms")/1($11.20)/(1 "kilograms") = ? Please notice how the kilograms cancel: (100cancel" kilograms")/1($11.20)/(1cancel" kilograms") = ?

The only remaining unit is dollars:

(100)/1($11.20)/(1) =$1120

This allows us to write equation [2]:

$10x +$11y + $12z =$1120 [2]

The phrase "If the same amounts of the two higher prices teas are used" allows us to write equation [3]:

$y = z$ [3]

Here are your 3 linear equations:

$x + y + z = 100$ kilograms [1]
$10x +$11y + $12z =$1120 [2]
$y = z$ [3]

Problem 14 setup:

Let x = the number of shirts
Let y = the number of ties

Selling everything for $10000 tells us that the right side of the equation [1] is$10000. The prices $100 for 3 shirts and$20 per tie tells us that the coefficients are ($100)/3 for x and$20 for y:

($100)/3x +$20y = $10000 [1] For the second equation, multiply x by $\frac{1}{2}$, multiply y by $\frac{2}{3}$and make the right side$6000

($100)/3(1/2)x +$20(2/3)y = $6000 [2] Simplify equation [2]: ($50)/3x + ($40)/3y =$6000 [2]

Here are your two linear equations:

($100)/3x +$20y = $10000 [1] ($50)/3x + ($40)/3y =$6000 [2]