Question

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!

Answer 1

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 `1/2`, multiply y by `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]