Simultaneous Equations - MathsTips.com

Let us consider the equation $x-y=2$ in which $x$ and $y$ are both unknown. It is very clear that this equation will hold true, i.e., be satisfied for all values of $x$ and $y$ . We can safely say that this equation has an infinite number of solutions because any pair of numbers whose difference is 2 will satisfy the equation. For example, $x=5, y=3$ ; $x=3, y=1$ ; $x=4, y=2$ ; and so on. However, if, together with this equation we consider the equation $x+y=8$ then $x$ and $y$ must be such that their difference is 2 and their sum is 8. Thus, the two equations $x-y=2; x+y=8$ will both be satisfied by the same pair of values of $x$ and $y$ only when $x=5$ and $y=3$

Again, let us consider the following equations:

$x+y+z=6 \\ x-y+z=4 \\ x+y-z=2$

These equations will be satisfied by the same values of $x, y, z$ only when $x=3, y=1, z=2$ . These equations may be individually satisfied for all values of the unknown quantities, but there is only one set of values of $x, y, z$ which will satisfy all three equations simultaneously. Two or more equations which are all satisfied by a single set of values of the involved variables are called simultaneous equations.

Consider the two linear equations $3x+4y= 6$ and $8x+ 5y = 3$ . Each of these two equations contains two variables (namely $x$ and $y$ ) and together these equations are called simultaneous (linear) equations. These equations are linear simultaneous equations or simple simultaneous equations because the maximum power of the variables involved in them is 1.

To solve two simultaneous linear equations means to find the values of unknown variables $x$ and $y$ satisfying both the given conditions.

Methods of Solving Simultaneous Equations

The methods for solving simultaneous equations are as follows:

Method of Substitution
Method of Elimination
Method of Comparison

1. Method of Substitution

Steps:

From any of the given two equations; find the value of one variable in terms of other.
Substitute the value of the variable, obtained in step 1 above, in the other equation and solve it to get the value of one variable.
Substituting the value of the variable obtained in step 2 above, in the result of step 1, get the value of the remaining unknown variable.

Solve the following system of equations by the Method of Substitution

Example: Solve for $x$ and $y$

$2x+ 3y= 7$

$3x - y= 5$

Solution:

$2x+ 3y= 7$ …(1)

$3x - y= 5$ …(2)

Step 1: From the second equation we have,

$3x - y=5$

or, $3x - 5=y$

or, $y=3x - 5$ …(3)

Step 2: Substituting this value in the first equation we have,

$2x+ 3y = 7$

or, $2x+ 3(3x - 5) = 7$

or, $2x+ 9x- 15 = 7$

or, $11x = 22 \Rightarrow x=2$

Step 3: Hence from (3) we get,

$y=3 \times 2-5$

or, $y= 6-5 =1$

$\therefore$ the required solution is:

$x=2$

$y=1$

Example: Solve for $x$ and $y$

$x+ 11y =1$

$8x+ 13y = 2$

Solution:

$x+ 11y =1$ …(1)

$8x+ 13y = 2$ …(2)

Step 1: From the first equation we have,

$x + 11y = 1$

or, $x= 1-11y$ …(3)

Step 2: Substituting this value in the first equation we have,

$8x + 13y = 2$

or, $8(1-11y) + 13y = 2$

or, $8- 88y +13y = 2$

or, $-75y=-6$

or, $y= \dfrac{-6}{-75} = \dfrac{2}{25}$

Step 3: Hence from (3) we get,

$x = 1-11 \times \dfrac{2}{25} = \dfrac{25-22}{25} = \dfrac{3}{25}$

$\therefore$ the required solution is:

$x = \dfrac{3}{25}$

$y = \dfrac{2}{25}$

2. Method of Elimination

Steps:

Multiply one or both of the equations by a suitable number or numbers so that either the coefficients of $x$ or the coefficients of $y$ in both the equations become numerically equal.
Add or subtract one equation from the other so that the terms with equal numerical coefficients cancel mutually and what remains is an equation containing either only $x$ or only $y$ .
Solve the resulting equation to find the value of the one of the unknowns.
Substitute this value in any of the two equations given and find the value of the other unknown.

Solve the following system of equations by the Method of Elimination

Example: Solve for $x$ and $y$

$3x - 4y = 10$

$5x - 3y = 24$

Solution:

$3x - 4y = 10$ …(1)

$5x - 3y = 24$ …(2)

Step 1: We multiply equation (1) by 5 and equation (2) by 3.

The resulting equations are:

$15x - 20y = 50$ …(3)

$15x - 9y = 72$ …(4)

Step 2: Subtracting equation (4) from equation (3) we have,

$15x-15x-20y+9y=50-72$

or, $-11y=-22$

Step3: $y=\dfrac{-11}{-22}$

or, $y=2$

$\therefore$ $y = 2$

Step 4: Substituting $y = 2$ in equation (1) we get,

$3x - 4 \times 2 = 10$

or, $3x = 10+8$

or, $x = \dfrac{18}{3} = 6$

$\therefore$ the required solution is:

$x= 6$

$y=2$

Note: Here we have chosen the numbers to be multiplied such that the coefficients of $x$ become equal and we can thus eliminate $x$ and arrive at an equation only in $y$ as can be seen in steps 2 and 3. We could have chosen the numbers in such a manner that the coefficients of $y$ would have become equal. Then we would have eliminated $y$ and arrive at an equation only in $x$ . This is shown below:

$3x - 4y = 10$ …(1)

$5x - 3y = 24$ …(2)

We multiply equation (1) by 3 and equation (2) by 4.

The resulting equations are:

$9x-12y=30$ …(3)

$20x-12y=96$ …(4)

Subtracting equation (4) from equation (3) we have,

$9x-20x-12y+12y=30-96$

or, $-11x=-66$

or, $x=\dfrac{-66}{-11}=6$

Substituting $x=6$ in equation (1) we have,

$3 \times 6 -4y=10$

or, $-4y=10-18$

or, $y=\dfrac{-8}{-4}=2$

Hence, it can be safely concluded that, which variable we choose to eliminate will not alter the solution in any way.

Example: Solve for $x$ and $y$

$3x-4y=5$

$5x+2y=17$

Solution:

$3x-4y=5$ …(1)

$5x+2y=17$ …(2)

Step 1: We multiply equation (2) by 2.

The resulting equations are:

$10x+4y=34$ …(3)

$3x-4y=5$ …(1)

Step 2: Adding equations (3) and (1) we have,

$10x+3x+4x-4y=34+5$

or, $13x=39$

Step 3: $x=\dfrac{39}{13}$

$\therefore$ $x=3$

Step 4: Substituting this value of $x$ in equation (1) we have,

$3 \times 3-4y=5$

or, $-4y=5-9$

or, $y=\dfrac{-4}{-4}=1$

$\therefore$ the required solution is:

$x=3$

$y=1$

Note: It should be kept in mind that whatever method is undertaken for solving the pair of simultaneous equations the solution will remain the same, i.e., it is immaterial which method one chooses to solve a given pair of simultaneous equations.

The third method of solving simultaneous equations viz., Method of Comparison will be covered in the article Simultaneous Equations II.

3. Method of Comparison

Steps:

From the first equation find the value of any one unknown quantity in terms of the other.
From the second equation find the value of the same unknown quantity in terms of the other. For example, say we have a pair of simultaneous equations in $x$ and $y$ . From the first equation we express the value of $x$ in terms of $y$ . From the second equation also, we express the value of $x$ in terms of $y$ .
These two values so obtained are obviously equal because $x=x$ and $y=y$ for all values of $x$ and $y$ . Equate these two values to obtain an equation in one variable and solve for that variable.
Substituting this value of one of the variables in any one of the earlier equations, find out the value of the other variable.

Solve the following system of equations by the Method of Comparison

Example: Solve for $x$ and $y$

$6x-5y=11$

$2x+3y=27$

Solution:

$6x-5y=11$ …(1)

$2x+3y=27$ …(2)

Step 1: From the first equation we have,

$5y=6x-11$

$\therefore$ $y=\dfrac{6x-11}{5}$ …(3)

Step 2: From the second equation we have,

$3y=27-2x$

$\therefore$ $y=\dfrac{27-2x}{3}$ …(4)

Step 3: From (3) and (4) we have,

$\dfrac{6x-11}{5}=\dfrac{27-2x}{3}$

or, $3(6x-11)=5(27-2x)$

or, $18x-33=135-10x$

or, $28x=168$

$\therefore$ $x=\dfrac{168}{28}=6$

Step 4: Substituting this value of $x=6$ in equation (3) we have,

$y=\dfrac{6 \times 6-11}{5}$

or, $y=\dfrac{25}{5}=5$

$\therefore$ the required solution is:

$x=6$

$y=5$

Example: Solve for $x$ and $y$

$\dfrac{7+x}{5}-\dfrac{2x-y}{4}=3y-5$

$\dfrac{5y-7}{2}+\dfrac{4x-3}{6}=18-5x$

Solution:

$\dfrac{7+x}{5}-\dfrac{2x-y}{4}=3y-5$ …(1)

$\dfrac{5y-7}{2}+\dfrac{4x-3}{6}=18-5x$ …(2)

Multiplying both sides of the first equation by 20 we have,

$4(7+x)-5(2x-y)=20(3y-5)$

or, $28-6x+5y=60y-100$

$\therefore$ $55y+6x=128$ …(3)

Multiplying both sides of the second equation by 6 we have,

$3(5y-7)+(4x-3)=6(18-5x)$

or, $15y+4x-24=108-30x$

$\therefore$ $15y+34x=132$ …(4)

Step1: From equation (3) we have,

$55y=128-6x$

or, $y=\dfrac{128-6x}{55}$ …(5)

Step 2: From equation (4) we have,

$15y=132-34x$

or, $y=\dfrac{132-34x}{15}$ …(6)

Step 3: From (5) and (6) we have,

$\dfrac{128-6x}{55}=\dfrac{132-34x}{15}$

or, $\dfrac{64-3x}{11}=\dfrac{66-17x}{3}$ [Multiplying both sides by $\dfrac{5}{2}$ ]

or, $3(64-3x)=11(66-17x)$

or, $192-9x=726-187x$

or, $178x=534$

$\therefore$ $x=3$

Step 4: Substituting this value of $x=3$ in equation (5) we have,

$y=\dfrac{128-6 \times 3}{55}$

or, $y=\dfrac{128-18}{55}$

or, $y=\dfrac{110}{55}=2$

$\therefore$ the required solution is:

$x=3$

$y=2$

Example: Solve for $x$ and $y$

$65x - 33y = 97$

$33x - 65y = 1$

Solution:

In this example, the coefficient of $x$ in the first equation is numerically equal to the coefficient of $y$ in second equation and the coefficient of $y$ in the first equation is numerically equal to the coefficient of $x$ in the second equation.

Such equations are solved by the method, given below:

$65x - 33y = 97$ …(1)

$33x - 65y = 1$ …(2)

On adding (1) and (2), we get:

$98x - 98y = 98$

or, $x-y=1$ …(3) (dividing each term by 98)

On subtracting (2) from (1), we get:

$32x + 32y = 96$

Or, $x+ y = 3$ …(4) (dividing each term by 32)

Now on solving equations (3) and (4), we get:

$x= 2$

$y= 1$

Problems leading to Simultaneous Equations

To solve a problem on simultaneous equations, adopt the following steps:

Assume the two variables (unknowns) as $x$ and $y$ .
According to the problem, set up two equations in terms of $x$ and $y$ .
Solve the pair of simultaneous equations by any of the methods that have been explained in this article and the other article on simultaneous equations.

Example: The sum of the digits of a two-digit number is 7. If the digits are reversed, the new number is increased by 3 and equals 4 times the original number. Find the original number.

Solution:

Let $x$ be the digit at ten’s place and $y$ be the digit at unit’s place.

$\therefore$ the number $= 10x +y$ and the sum of the digits $= x+y$ .

On reversing the digits, the number becomes $10y+x$

According to the problem:

$x+y = 7$

$10y+ x +3 = 4(10x + y)$

On solving we get: $x= 1, y=6$

[Here any one of the methods illustrated may be used to solve these equations]

$\therefore$ the required number $= 10x +y = 10 \times 1 +6 = 16$

Example: A and B each have certain number of oranges. A says to B. “If you give me 10 of your mangoes, I will have twice the number of oranges left with you”. B replies, “If you give me 10 of your oranges, I will have the same number of oranges as left with you”. Find the number of oranges with A and B respectively.

Solution:

Let A has $x$ number of oranges and B has $y$ number of oranges.

In 1^st case (if B gives oranges to A):

$x+ 10 = 2(y- 10)$

or, $x -2y = -30$ …(1)

In 2^nd case (If A gives 10 oranges to B)

$y + 10 = x - 10$

or, $x - y = 20$ …(2)

On solving equations (1) and (2), we get:

$x=70$ and $y = 50$ .

$\therefore$ A has 70 oranges and B has 50 oranges.

Exercise

Solve the following pairs of linear (simultaneous) equations by the method of comparison:
1. $\dfrac{1}{5} (x-2)=\dfrac{1}{4} (1-y); 26x+3y+4=0$
2. $x+ y = 2xy; x- y = 6xy$
3. $5x-3y=9; 5y+2x=16$
4. $y(3+x)=x(7+y); 4x+9=5y-14$
5. $3x-7y=7; 11x+5y=87$
The sides of an equilateral triangle are given by $x+3y, 3x+2y-2$ and $4x+ \dfrac{1}{2}y +1$ respectively. Find the lengths of the sides of the triangle.
A and B both have pencils. If A gives 10 pencils to B, then B will have twice as many as A and if B gives 10 pencils to A, then they will have the same number of pencils. How many pencils does each have?
In an examination the ratio of passes to failures was 4:1. Had 30 less appeared and 20 less passed, the ratio of passes to failures would have been 5:1. Find the number of students who appeared for the examination.
A farmer wishing to purchase a certain number of sheep found that if they cost him Rs 42 a head, he would not have money enough by Rs 28, but if they cost him Rs 40 a head, he would then have Rs 40 more than what he required. Find the number of sheep and the money which he had?
A and B are friends and their ages differ by 2 years. A’s father D is twice as old as A and B is twice as old as his sister C. the ages of D and C differ by 40 years. Find the ages of A and B.
A certain number between 10 and 100 is eight times the sum of its digits and if 45 be subtracted from it, the digits will be reversed. Find the number.
Two men and 5 boys together can finish a piece of work in 4 days while 3 men and 6 boys can finish it in 3 days. Find the time taken:
1. By one man alone to finish the work
2. By one boy alone to finish the work.
What fraction is that which, if 1 be added to the numerator, becomes 1, and, if 1 be added to the denominator becomes $\dfrac{1}{2}$ ?
A and B lay a wager of Rs 10. If A loses, he will have Rs 25 less than twice as much money as B will then have, but if B loses, he will have $\dfrac{5}{17}$ th of what A will then have. Find how much money each of them has.
[Hint: A wager of Rs 10 means, if A loses not only he himself have Rs 10 less than what he originally had, but simultaneously B will receive this Rs 10 from A and hence B will have Rs 10 more than what he originally had. Let A and B both originally have $Rs \: x$ and $Rs \: y$ respectively.]
Solve the following pairs of linear (simultaneous) equations by the method of substitution:
1. $y= 4x - 7; 16x - 5y = 25$
2. $15x - 8y = 29; 17x+ 12y = 75$
3. $7x-5y=11; 3x+2y=13$
4. $8x-9y=20; 7x-10y=9$
5. $2x+3y=32; 11y-9x=3$
Solve the following pairs of linear (simultaneous) equations by the method of elimination:
1. $ax + by = a- b; bx -ay = a+ b$
2. $x+4y=14; 7x-3y=5$
3. $5x-8y=9; 13x+7y=79$
4. $x+ay=b; ax-by=c$
5. $ax+by=c; a^2x+b^2y=c^2$

Comments

Aditya Gor says

October 1, 2014 at 4:41 am

How to solve this sum :

the sum of the digits of the two digit number is 9. The number obtained by interchanging the digits exceeds the original number by 45. Find the original number.

James Houston Carruthers says

December 17, 2021 at 5:47 am

solving simultaneous equations

Step 2: Subtracting equation (4) from equation (3) we have,

15x-15x-20y+9y=50-72

or, -11y=-22

Step3: y=(-11) / (-22)
or, y=2

**************the error is in Step3 ***********************************
**************which mysteriously still gives the correct answer ***
**************on the next line*****************************************

Methods of Solving Simultaneous Equations

1. Method of Substitution

Solve the following system of equations by the Method of Substitution

2. Method of Elimination

Solve the following system of equations by the Method of Elimination

3. Method of Comparison

Solve the following system of equations by the Method of Comparison

Problems leading to Simultaneous Equations

Exercise

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