Examples of simultaneous equations word problems (step by step)

Examples of simultaneous equations word problems and how to solve very fast

Simultaneous equations are two or more algebraic equations that share variables, e.g. x and y. In today’s article, I will show you different examples of how to solve any given type of simultaneous equations word problems step by step very, very fast.

We can consider each equation in a simultaneous equation as a function which when displayed graphically may intersect at a specific point. This point of intersection gives the solution to the simultaneous equations.

Examples of simultaneous equations word problems

Some examples of word problems of simultaneous equations are:

  1. A jet weighs 225Kg when it’s half full of passengers and 380Kg when it’s three quarters full of passengers. Calculate the weight of the jet when it’s empty
  2. In Obafemi Awolowo University, two students went into the bookshop and bought the books. One student bought 5 different Chemistry text books and 3 different Maths text books at a price of N150 while the second student bought 2 different Chemistry text books and 1 Maths text books at a price of N90. Calculate the price of Chemistry text books and that of Maths text books.

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Solution to #1

Solutions of number one of the example of simultaneous equation words problems listed above are as follows

Let J and P represent the weight of jet and passengers respectively.

Now, at the time when the jet is half full of passengers, the best equation we can use is J + 1/2(P) = 225Kg when the best equation we can use when the jet contains three quarters of passengers is J + 3/4(P) = 380Kg.

Meanwhile, the weight of the jet when it’s not filled with anything is unknown and we’re being asked to find it out from the example of simultaneous word problem question given above.

Let’s start simultaneously solving this question.

Therefore,

J + 1/2(P) = 225Kg …. equation one

J + 3/4(P) = 380Kg …. equation two

Now, let’s solve it simultaneously using elimination method such that equation two substract equation one.

3/4(P) – 1/2(P) = 380Kg – 225Kg

2P/4 = 155Kg

Now, let’s cross multiply

4(155Kg) = 2P

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620Kg = 2P

P = 310Kg.

Since we have just known the weight of passengers only as 310Kg, now let’s substitute the value of the weight of passengers into equation one, J + 1/2(P) = 225Kg.

J + 1/2(310Kg) = 225Kg

J + 155Kg = 225Kg

J = 225Kg – 155Kg

J = 70Kg

From the solution above, it’s now known that the weight of jet only (J) is 70Kg while that of passengers only (P) is 3310Kg.

To check whether this solution is correct or not, just simply add weight of jet and weight of passengers altogether and you will have 380Kg, i.e, 70Kg + 310Kg = 380Kg which is the weight of the jet when it’s three quarters filled with passengers.

Solution to #2

Let C and M represent the price of Chemistry text books and Maths text books,. respectively.

Therefore, from the question given in number above, we can deduce that:

5C+ 3M = 150 … equation one

2C + M = 90 … equation two.

From equation two, let us make M the subject of formula, hence:

2C + M = 90

M = 90 – 2C … equation three.

Now, let’s substitute the value of M in equation one

Therefore, 5C + 3(90 – 2C) = 150

5C + 270 – 6C = 150

C = 120

Now, let’s substitute the value of C, 120 into equation three, M = 90 – 2C

M = 90 – 2(120)

M = 90 – 240

M = 30

Therefore, the price of Chemistry text book is N120 and that of Maths is N30.

Conclusion

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I am Abdulrazaq, a student and a type of person who love guiding people with any important thing I know. I established this blog in order to show people what is what and prevent them from making avoidable mistakes.

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