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May 20, 202239 Insightful Publication

**Word Problems on Simultaneous Linear Equations: **When we have to find the values of two unknown quantities in a problem, we take the two unknown quantities to be x, y, or any two other algebraic symbols. Then, based on the stated condition or conditions, we build the equation and solve the two simultaneous equations to determine the values of the two unknown numbers. As a result, we can solve the problem. Simultaneous equations are ones that need simultaneous solutions. Word problems, also known as applied problems, are situations in which unknowns are described using words.

For success with word (or applied) difficulties, practice and mastery of certain basic translation processes are essential. Solving word problems can be done in two ways. To begin, translate the problems’ wording into algebraic equations. The second step is to solve the resulting equations. We will go over word problems on simultaneous linear equations in-depth in this article.

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Mathematical issues rarely reveal themselves as \(2+3\) or \(6-4\) in real life. They are, in fact, a little more complicated than we believe. Authors of mathematics curricula occasionally use word problems to assist students to comprehend how their work relates to the actual world.

Word problems frequently represent mathematics as it occurs in the real world. The difficulty of word problems can range from easy to complicated.

Here are a few examples to get you some ideas:

- Keerthi had \(5\) chocolates. Her mother gave her \(7\) more chocolates. Now, how many chocolates does Keerthi have altogether?
- There were \(18\) pens and \(9\) erasers. How many more pens than erasers are there?
- Kriti has one dozen bananas. Her friends ate \(4\) for breakfast. Now, how many bananas are left with Kriti?
- There are \(45\) mangoes. Jyoti, Priya, and Preethu want to eat them in equal shares. How many mangoes will each of the friends get?

As you can notice, word problems involve addition, subtraction, multiplication, division or even multiple operations.

Simultaneous linear equations in two variables depict real-life situations by involving two unknown values.

It aids in establishing a link between quantities, pricing, speed, time, distance, and other factors, which leads to a better knowledge of the issues.

Simultaneous linear equations are used in our daily lives without us even realising them.

No approach works in all instances due to the great range of word (or application) issues. The following broad ideas, however, might be helpful:

(i) Carefully read and re-read the problem statement to discover what amounts must be found.

(ii) Use letters to represent the unknown quantities.

(iii) Write equations by determining whether expressions are equivalent.

(iv) Solve the resulting equations.

Let us take a mathematical problem to indicate the necessary steps for forming simultaneous equations:

In a stationery shop, the cost of \(3\) pencil cutters exceeds the price of \(2\) pens by \(â‚¹2.\) Also, the total price of \(7\) pencil cutters and \(3\) pens is \(â‚¹43.\)

Follow the steps of instructions.**Step I:** Identify the unknown variables; assume one of them as \(x\) and the other as \(y\)

Here two unknown quantities (variables) are:

Price of each pencil cutter\(=â‚¹x\)

Price of each pen\(=â‚¹y\)**Step II:** Identify the relationship between the unknown quantities.

Price of \(3\) pencil cutters\(=â‚¹3x\)

Price of \(2\) pens\(=â‚¹2y\)

Therefore, the first condition gives: \(3x-2y=2\)**Step III:** Express the conditions of the problem in terms of \(x\) and \(y\)

Again price of \(7\) pencil cutters\(=â‚¹7x\)

Price of \(3\) pens\(=â‚¹3y\)

Therefore, the second condition gives: \(7x+3y=43\)

Simultaneous equations formed from the problem is

\(3x – 2y = 2 \ldots \ldots \ldots {\rm{(i)}}\)

\(7x – 3y = 43 \ldots \ldots \ldots {\rm{(ii)}}\)

Example: Twice one number minus three times a second equals \(2,\) and the sum of these numbers is \(11.\) Find the numbers.

Solution: Let the two numbers be \(x, y.\)

According to the problem,

\(3x – 3y = 2 \ldots ..{\rm{(i)}}\)

and \(x – y = 11 \ldots {\rm{(ii)}}\)

Multiplying both sides of \({\rm{(ii)}}\) by \(3,\) we get

\(3x – 3y = 33 \ldots ..{\rm{(iii)}}\)

On adding \({\rm{(i)}}\) and \({\rm{(iii)}},\) we get

\(5x=35â‡’x=7\)

Substituting this value of \(x\) in \({\rm{(ii)}},\) we get

\(7+y=11â‡’y=11-7â‡’y=4\)

Hence, the required numbers are \(7\) and \(4.\)

**Q.1: A man buys postage stamps of denominations** \(25\) **paise and** \(50\) **paise for** \(â‚¹10.\) **He buys** \(28\) **stamps in all. Find the number of** \(25\) **paise stamps bought by him.****Ans:** Let the number of \(25\) paise stamps be \(x,\) and the number of \(50\) paise stamps be \(y.\)

According to the problem,

\(x + y = 28 \ldots {\rm{(i)}}\)

and \(25x+50y=1000\) \([â‚¹10=1000]\) paise i.e.

\(x + 2y = 40 \ldots ({\rm{ii}})\)

Subtracting \({\rm{(i)}}\) from \({\rm{(ii)}},\) we get,

\(y=12\)

Substituting this value of \(y\) in \({\rm{(i)}},\) we get

\(x+12=28\)

\(â‡’x=28-12\)

\(â‡’x=16\)

Hence, the number of \(25\) paise stamps is \(16.\)

**Q.2: A chemist has one solution,** \(50\% \) **acid, and a second,** \(25\% \) **acid. How much of each should be mixed to make** \({\rm{10}}\,{\rm{litres}}\)** of a** \(40\% \) **acid solution?****Ans:** Let \(x\) litres of \(50\% \) and y litres of \(25\% \) acid solutions be mixed. Then, according to the given conditions, we have

\(x + y = 10 \ldots ({\rm{i}})\)

and \(25\% \) of \(x + 50\% \) of \(y = 40\% \) of \(10\)

\( \Rightarrow \frac{{50}}{{100}}x + \frac{{25}}{{100}}y = \frac{{40}}{{100}} \times 10\)

\( \Rightarrow \frac{1}{2}x + \frac{1}{4}y = 4\)

\( \Rightarrow 2x + y = 16 \ldots {\rm{(ii)}}\)

Subtracting equation \({\rm{(i)}}\) from equation \({\rm{(ii)}},\) we get \(x=6\)

Substituting this value of \(x\) in equation \({\rm{(i)}},\) we get

\(6+y=10â‡’y=4\)

Hence, \(6\,{\rm{litres}}\) of \(50\% \) and \(4\,{\rm{litres}}\) of \(25\% \) acid solutions be mixed to get \(10\,{\rm{litres}}\) of \(40\% \) acid solution.

**Q.3: A two-digit number is seven times the sum of its digits. The number formed by reversing the digits is** \(18\) **less than the original number. Find the number.****Ans:** Let \(x\) be the digit at ten’s place, and y be the digit at unit’s place.

Then the number is \(10x+y.\)

According to the first condition of the problem,

\(10x+y=7(x+y)\)

\(â‡’10x+y=7x+7y\)

\(â‡’3x=6y\)

\( \Rightarrow x + 2y \ldots ({\rm{i}})\)

The number formed by reversing the digits is \(10y+x.\)

According to the second condition of the problem,

\(10y+x=(10x+y)-18\)

\(â‡’10y-y=10x-x-18\)

\(â‡’9y=9x-18\)

\(â‡’y=x-2\)

\(â‡’y=2y-2\) [Using \(({\rm{i}})\)]

\(â‡’y=2\)

From \(({\rm{i}}),\) \(x=2Ã—2=4\)

Hence, the required number is \(42.\)

**Q.4: The number of heads and legs of deer and human visitors at a deer park was counted at one point, and it was found that there were** \(41\) ** heads and** \(136\)

As a deer and a human each has one head, we get

\(x + y = 41 \ldots ({\rm{i}})\)

As a deer has \(4\) legs and a human has \(2\) legs, we get

\(4x+2y=136\)

\( \Rightarrow 2x + y = 68 \ldots ({\rm{ii}})\)

Subtracting \({\rm{(i)}}\) from \(({\rm{ii}}),\) we get

\(x=27\)

Substituting \(x=27\) in \({\rm{(i)}},\) we get

\(27+y=41\)

\(â‡’y=14\)

Hence, there were \(27\) deer and \(14\) human visitors in the park.

**Q.5: Six years after a woman’s age will be three times her daughter’s age, and three years ago, she was nine times as old as her daughter. Find their present ages.****Ans:** Let the present age of the woman be \(x\) years, and the present age of her daughter be \(y\) years.

\(6\) years after, their ages will be \((x+6)\) years and \((y+6)\) years.

According to the problem,

\(x+6=3(y+6)\)

\(â‡’x+6=3y+18\)

\( \Rightarrow x + 3y = 12 \ldots ({\rm{i}})\)

3 years ago, their ages were \((x-3)\) years and \((y-3)\) years.

According to the problem,

\(x-3=9(y-3)\)

\(â‡’x-3=9y-27\)

\( \Rightarrow x – 9y = – 24 \ldots ({\rm{ii}})\)

Subtracting \({\rm{(ii)}}\) from \({\rm{(i)}},\) we get

\(6y=36â‡’y=6\)

Substituting this value of \(y\) in \({\rm{(i)}},\) we get

\(x-3Ã—6=12\)

\(â‡’x-18=12\)

\(â‡’x=12+18=30\)

Hence, the present age of the woman is \(30\) years, and that of her daughter is \(6\) years.

In this article, we have discussed word problems and how to solve a word problem. Simultaneous equations are those that must be solved at the same time. Word problems or applied problems are real-life situations involving unknowns that are described in words. Also, we covered simultaneous linear equations and the necessary steps to solve word problems on simultaneous linear equations. And at last, we had some solved word problems on simultaneous linear equations.

**Q.1: Why is it called a simultaneous equation?****Ans:** Sometimes, you’ll come across two or more unknown quantities, as well as two or more equations that relate to them. Simultaneous equations are what they’re called. As a result, the only pair of numbers that answer both equations simultaneously may be found in this manner. As a result, these are referred to as simultaneous equations.

**Q.2: What is the difference between linear and simultaneous equations?****Ans:** A linear equation is one in which all the terms are either constants or the products of constants. Simultaneous equations are a collection of equations with several variables.

**Q.3: How do you create a simultaneous equation from a word problem?****Ans:** Step I: Identify the unknown variables; assume one of them as \(x\) and the other as \(y\)

Step II: Identify the relationship between the unknown quantities.

Step III: Express the conditions of the problem in terms of \(x\) and \(y\)

Step IV: Solve for the values of \(x\) and \(y\)

**Q.4: What are linear equations in two variables?****Ans:** An equation is said to be a linear equation in two variables if it is written in the form of \(Ax+By+C=0,\) where \(A, B\& C\) are real numbers and the coefficients of \(x\) and \(y,\) and \(aâ‰ 0,bâ‰ 0\)

For example, \(6x+4y=3\) and \(-3x+5y=12\) are linear equations in two variables.

**Q.5: How many ways can you solve simultaneous equations?****Ans:** You can solve for both unknowns if you have two separate equations with the same two unknowns in each. The elimination, substitution, and graphical methods are the three most popular approaches for solving problems.

Learn to Algebraically Solve Simultaneous Linear Equations here

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