ICSE Class 8 Maths Algebra Chapter 09 Simultaneous Linear Equatins PDF Download

Simultaneous Linear Equations

Linear Equation In Two Variables

A linear equation in two variables (say, x and y) contains the variables in the first degree and in separate terms. The general form of such an equation is \(ax + by + c = 0\), where a, b and c are real numbers and a and b are nonzero numbers.

Examples: \(2x - 5y = 7\) and \(5x + \frac{y}{2} + 3 = 0\) are linear equations in x and y. However, \(5xy = 9\) is not a linear equation because x and y are not contained in separate terms.

Simultaneous Linear Equations

If two linear equations in x and y are satisfied by the same values of x and y then the equations are called simultaneous linear equations. The general form of such equations is \(ax + by + c = 0\) and \(px + qy + r = 0\).

Example: The equations \(x + 2y = 5\) and \(2x + y = 4\) are satisfied by the values \(x = 1, y = 2\). Therefore, \(x + 2y = 5\) and \(2x + y = 4\) are simultaneous linear equations and their solution is \(x = 1, y = 2\).

To find a solution to simultaneous linear equations, we must find a pair of values of the variables that satisfy both the equations. There are two ways of doing this.

By substitution or By elimination

Substitution Method

This method involves taking the following steps.

Step 1: Using one of the equations write y in terms of x (or x in terms of y) and the constant.

Step 2: Substitute the expression for y (or x) in the second equation.

Step 3: Solve the resulting linear equation in x (or y).

Step 4: Substitute the value of x (or y) in either of the equations.

Step 5: Solve the resulting linear equation in y (or x).

Step 6: Verify the correctness of the solution by substituting the values of x and y in the given equations.

Example

Solve the equations \(5x + y = 10\) and \(14x + 3y = 18\).

Given, \(5x + y = 10\) ... (1)

\(14x + 3y = 18\) ... (2)

From the equation (1), \(5x + y = 10\) or \(y = 10 - 5x\).

Substituting the expression \(10 - 5x\) for y in equation (2),

\(14x + 3(10 - 5x) = 18\) or \(14x + 30 - 15x = 18\) or \(-x + 12 = 0\) or \(x = 12\).

Substituting \(x = 12\) in the equation (2),

\(14 \times 12 + 3y = 18\) or \(3y = 18 - 168 = -150\) or \(y = -50\)

Hence, the solution is \(x = 12, y = -50\).

Teacher's Note

When solving word problems about combining items or mixing quantities, we use simultaneous equations - like figuring out the cost of apples and oranges when we know the total price of different combinations.

Elimination Method

This method is also called the addition-subtraction method.

Step 1: Decide which variable will be easier to eliminate. Try to avoid fractions.

Step 2: Multiply one or both the equations by suitable numbers to ensure that the coefficients of the variable to be eliminated are the same in both the equations.

Step 3: Add or subtract the resulting equations to eliminate the variable.

Step 4: Solve the resulting equation in one variable.

Step 5: Substitute the value of the variable obtained in Step 4 in either of the given equations.

Step 6: Solve the resulting equation.

Step 7: Verify the correctness of the solution by substituting the values of the variables in the given equations.

Example

Solve the equations \(5x + 3y = 7\) and \(2x + 5y = 1\).

Given, \(5x + 3y = 7\) ... (1)

\(2x + 5y = 1\) ... (2)

To eliminate x, let us multiply the equation (1) by 2 and the equation (2) by 5. Thus, \(10x + 6y = 14\) ... (3)

\(10x + 25y = 5\) ... (4)

Subtracting the equation (4) from the equation (3),

\(-19y = 9\) or \(y = \frac{9}{-19} = -\frac{9}{19}\).

Substituting \(y = -\frac{9}{19}\) in the equation (1),

\(5x + 3 \times \frac{-9}{19} = 7\) or \(5x = 7 + \frac{27}{19} = \frac{133 + 27}{19} = \frac{160}{19}\)

\(\therefore x = \frac{1}{5} \times \frac{160}{19} = \frac{32}{19}\)

\(\therefore\) the solution is \(x = \frac{32}{19}, y = -\frac{9}{19}\).

Substitute \(x = \frac{32}{19}, y = -\frac{9}{19}\) in the equation (2),

\(2 \times \frac{32}{19} + 5 \times \frac{-9}{19} = 1\) or \(\frac{64}{19} - \frac{45}{19} = 1\) or \(\frac{19}{19} = 1\)

which is true. Hence, the solution is correct.

Teacher's Note

The elimination method is like balancing a seesaw - if we add or remove the same weight from both sides, the balance is maintained, which helps us solve for one variable at a time.

Solved Examples

Example 1

Solve \(3y - 2x = 1, 3x + 4y = 24\) by the substitution method.

Given, \(3y - 2x = 1\) ... (1)

\(3x + 4y = 24\) ... (2)

Let us solve the equation (1) for x.

\(3y - 2x = 1\) or \(2x = 3y - 1\) or \(x = \frac{3y - 1}{2}\).

Substituting this value of x in the equation (2),

\(3 \times \frac{3y - 1}{2} + 4y = 24\).

Multiplying both sides by 2,

\(3(3y - 1) + 8y = 48\) or \(9y - 3 + 8y = 48\) or \(17y = 48 + 3 = 51\).

\(\therefore y = \frac{51}{17} = 3\).

Substituting the value of y in the equation (1),

\(3 \times 3 - 2x = 1\) or \(9 - 2x = 1\) or \(2x = 9 - 1 = 8\) or \(x = 4\).

Hence, the solution is \(x = 4, y = 3\).

Example 2

Solve \(8a + 5b = 9\) and \(3a + 2b = 4\).

Given, \(8a + 5b = 9\) ... (1)

\(3a + 2b = 4\) ... (2)

To eliminate a, let us multiply the equation (1) by 3 and the equation (2) by 8. Thus, \(24a + 15b = 27\) ... (3)

\(24a + 16b = 32\) ... (4)

Subtracting the equation (3) from the equation (4),

\(b = 32 - 27 = 5\).

Substituting the value of b in the equation (1),

\(8a + 5 \times 5 = 9\) or \(8a = 9 - 25 = -16\) or \(a = \frac{-16}{8} = -2\).

Hence, the solution is \(a = -2, b = 5\).

Example 3

Solve \(7x + 3(y - 3) = 5(x + y)\) and \(7(x - 1) - 6y = 5(x - y)\).

Given, \(7x + 3(y - 3) = 5(x + y)\) ... (1)

\(7(x - 1) - 6y = 5(x - y)\) ... (2)

Simplifying the two equations,

\(7x + 3y - 9 = 5x + 5y\) or \(2x - 2y = 9\) ... (3)

\(7x - 7 - 6y = 5x - 5y\) or \(2x - y = 7\) ... (4)

Subtracting (3) from (4), \(y = 7 - 9 = -2\).

Substituting \(y = -2\) in (4), \(2x - y = 7\) or \(2x - (-2) = 7\) or \(2x = 5\) or \(x = \frac{5}{2}\).

Hence, the solution is \(x = \frac{5}{2}\) and \(y = -2\).

Teacher's Note

Sometimes equations are given in expanded form and need to be simplified first - like simplifying a recipe before measuring ingredients for cooking.

Example 4

Solve \(\frac{2}{x} - y = 2\) and \(\frac{3}{x} + 2y = 10\).

Given, \(\frac{2}{x} - y = 2\) and \(\frac{3}{x} + 2y = 10\).

Substituting \(\frac{1}{x} = z\), the given equations become

\(2z - y = 2\) ... (1)

and \(3z + 2y = 10\) ... (2)

Multiplying (1) by 2, \(4z - 2y = 4\) ... (3)

Adding (2) and (3), \(3z + 4z = 10 + 4\) or \(7z = 14\).

\(\therefore z = \frac{14}{7} = 2\), that is, \(\frac{1}{x} = 2\) or \(x = \frac{1}{2}\).

Substituting the value of z in (1), \(2 \times 2 - y = 2\) or \(y = 4 - 2 = 2\).

Hence, the solution is \(x = \frac{1}{2}, y = 2\).

Example 5

Solve \(\frac{2}{x} + \frac{3}{y} = -1\) and \(\frac{3}{x} + \frac{5}{y} = -2\).

Let \(\frac{1}{x} = p\) and \(\frac{1}{y} = q\).

Then \(2p + 3q = -1\) ... (1)

and \(3p + 5q = -2\) ... (2)

Multiplying (1) by 3 and (2) by 2,

\(6p + 9q = -3\) ... (3)

\(6p + 10q = -4\) ... (4)

Subtracting (3) from (4), \(q = -4 + 3 = -1\) or \(\frac{1}{y} = -1\) or \(y = -1\).

Substituting the value of q in (1),

\(2p + 3 \times (-1) = -1\) or \(2p - 3 = -1\) or \(2p = 3 - 1 = 2\)

or \(p = 1\) or \(\frac{1}{x} = 1\) or \(x = 1\).

Hence, the solution is \(x = 1, y = -1\).

Teacher's Note

When fractions appear in equations with reciprocals, we can use substitution to turn them into simpler linear forms - much like unpacking nested boxes to find what is inside.

Remember These

An equation of the form \(ax + by + c = 0\), where a, b and c are real numbers and a and b are nonzero numbers, is called a linear equation in two variables.

When two linear equations are satisfied by the same values of the two variables, the equations are called simultaneous equations.

There are two ways of solving simultaneous equations - by substitution and by elimination.

The substitution method essentially consists of using one of the equations to express one variable in terms of the other and then substituting this expression for the variable in the second equation.

The elimination method essentially consists of multiplying one or both the equations by suitable numbers so that one of the variables can be eliminated by adding or subtracting the resulting equations.

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