Exercise 5
(In this exercise use the information given in each case to form simultaneous
equations. Use substitution or elimination method to
solve them.)
1.
Danny had Ksh. 10 500 to spend on clothes. If he buys two shirts and
three pairs of trousers, he spends all the money. If he buys
three similar
shirts and 2 similar pairs of trousers, he would have Ksh. 1000 left. Find
the cost of one shirt and one pair of trousers.



Answer :

Alright, let's break down the problem step-by-step using the information provided.

1. Form the equations based on the given information:

Let:
- [tex]\( s \)[/tex] be the cost of one shirt in Ksh.
- [tex]\( t \)[/tex] be the cost of one pair of trousers in Ksh.

First scenario:
- Danny buys 2 shirts and 3 pairs of trousers for a total of Ksh. 10,500.
This can be expressed mathematically as:
[tex]\[ 2s + 3t = 10500 \tag{1} \][/tex]

Second scenario:
- Danny buys 3 shirts and 2 pairs of trousers, and he is left with Ksh. 1000.
Since he had originally Ksh. 10,500, the amount he spends is:
[tex]\[ 10500 - 1000 = 9500 \text{ Ksh} \][/tex]
This can be expressed mathematically as:
[tex]\[ 3s + 2t = 9500 \tag{2} \][/tex]

So we have the following system of equations:
[tex]\[ \begin{cases} 2s + 3t = 10500 \tag{1} \\ 3s + 2t = 9500 \tag{2} \end{cases} \][/tex]

2. Solve the system of simultaneous equations using the elimination method:

First, we multiply equation (1) by 3 and equation (2) by 2, in order to eliminate one of the variables when we subtract the equations:

[tex]\[ \begin{cases} 3(2s + 3t) = 3(10500) \implies 6s + 9t = 31500 \tag{3} \\ 2(3s + 2t) = 2(9500) \implies 6s + 4t = 19000 \tag{4} \end{cases} \][/tex]

Now, subtract equation (4) from equation (3):

[tex]\[ (6s + 9t) - (6s + 4t) = 31500 - 19000 \][/tex]

Simplify:

[tex]\[ 5t = 12500 \][/tex]

Divide both sides by 5:

[tex]\[ t = 2500 \][/tex]

So the cost of one pair of trousers is Ksh. 2500.

3. Find the cost of one shirt by substituting [tex]\( t = 2500 \)[/tex] into one of the original equations:

Substitute [tex]\( t = 2500 \)[/tex] into equation (1):

[tex]\[ 2s + 3(2500) = 10500 \][/tex]

Simplify:

[tex]\[ 2s + 7500 = 10500 \][/tex]

Subtract 7500 from both sides:

[tex]\[ 2s = 3000 \][/tex]

Divide by 2:

[tex]\[ s = 1500 \][/tex]

So the cost of one shirt is Ksh. 1500.

4. Verify the solution by substituting both [tex]\( s \)[/tex] and [tex]\( t \)[/tex] back into the second equation:

Using the values [tex]\( s = 1500 \)[/tex] and [tex]\( t = 2500 \)[/tex] in equation (2):

[tex]\[ 3(1500) + 2(2500) = 4500 + 5000 = 9500 \][/tex]

The solution satisfies both equations, confirming our values.

Answer:

- The cost of one shirt is Ksh. 1500.
- The cost of one pair of trousers is Ksh. 2500.