Many problems in everyday life can be solved mathematically through a system of linear equations. Usually these questions are analyzed with the following steps:
(1) Determine the objects that will be used as variables
(2) Look for relationships between variables in the form of a system of linear equations
(3) Determine the solution to the system of linear equations using the substitution or elimination method.
For more details will be described in the following example:
01. If two numbers have the sum of 45 and the difference is 5, then determine the two numbers!
Answer
Suppose the two numbers are x and y, then:
x + y = 45 ……………………………. (1)
x – y = 5 ………………………………(2)
Both equations are eliminated
The value of y is eliminated to (1) x + y = 45
x + 20 = 45
x = 25. So the two numbers are 25 and 20
02. If the numerator of a fraction is increased by 1 and the denominator is reduced by 3, the quotient equals 1/2, If the numerator is neither added nor subtracted, but the denominator is added by 1, the result is equal to 1/5. Determine the fraction
Answer
Answer
Suppose x = the price of one notebook
y = the price of one pencil
then 4x + y = 5600 …………………………………..…. (1)
5x + 3y = 8400 …………………………………………. (2)
Both equations are eliminated
(1) 4x + y = 5600
4(1200) + y = 5600
4800 + y = 5600
y = 5600 – 4800
y = 800
So the price of one notebook is Rp. 1,200 and one pencil Rp. 800
04. In a theater there are 400 spectators. The price of each ticket for class II is Rp. 5000, – and for class I Rp. 7,000,- The proceeds from the sale of tickets amounted to Rp. 2,300,000. how many spectators bought class II tickets and how many spectators bought class I tickets?
Answer
Suppose x = Number of class I viewers
y = Number of class II viewers
then x + y = 400 ……..……………………………………(1)
7000x + 5000y = 2,300,000 …… …………………… (2)
Both equations are eliminated
(1) x + y = 400
150 + y = 400
y = 400 – 150
y = 250 people
So the number of spectators who bought class I tickets was 150 people
The number of spectators who bought tickets for class II was 250 people
05. Five years ago Ali’s age was three times Wati’s age. Five years later Ali’s age became twice the age of Wati. How old are Ali and Wati now?
Answer
Suppose A = Ali’s current age
B = Wati’s current age
Then: A – 5 = 3(W – 5)
A – 5 = 3W – 15
A – 3W = 5 – 15
A – 3W = –10 ……………… …………… (1)
A + 5 = 2(W + 5)
A + 5 = 2W + 10
A – 2W = –5 + 10
A – 2W = 5 ……………………………… (2)
Both equations are eliminated A
(1) A – 3W = –10
A – 3(15) = –10
A – 45 = –10
A = 35 years
So Ali’s age is now 35 years and Wati 15 years
06. Six years ago the sum of the ages of father and mother was the same with 54 years. Now the age of the father is six fifths of the age of the mother. Determine each father’s age and mother’s age five years from now!
Answer
Suppose A = Father’s present age
B = Mother’s present age
Then: (A – 6) + (B – 6) = 54
A – 6 + B – 6 = 54
A + B – 12 = 54
A + B = 66 …… …….………………………………. (1)
(1) A + B = 66
A + 30 = 66
A = 66 – 30 = 36 years
So Father’s age 5 years from now is = 36 + 5 = 41 years
Mother’s age 5 years from now is = 30 + 5 = 35 years