ICSE Class 9 Maths Chapter 06 Simultaneous Linear Equations Including Problems

Simultaneous (Linear) Equations [Including Problems]

Introduction

An equation of the form \(ax + by + c = 0\) is called a linear equation in which \(a\), \(b\) and \(c\) are constants (real numbers) and \(x\) and \(y\) are variables each with degree 1 (one).

Consider the two linear equations: \(3x + 4y = 6\) and \(8x + 5y = 3\). These two equations contain same two variables (\(x\) and \(y\) in this case). Together such equations are called simultaneous (linear) equations.

Consider the simultaneous linear equations \(2x - y = 1\) and \(3x + y = 14\).

If \(x = 3\) and \(y = 5\)

\(2x - y = 1\) \(\Rightarrow\) \(2 \times 3 - 5 = 1\) \(\Rightarrow\) \(1 = 1\)

\(3x + y = 14\) \(\Rightarrow\) \(3 \times 3 + 5 = 14\) \(\Rightarrow\) \(14 = 14\)

Since, \(x = 3\) and \(y = 5\) satisfy both the equations \(2x - y = 1\) and \(3x + y = 14\). Therefore, \(x = 3\) and \(y = 5\) is the solution of simultaneous linear equations under consideration.

Teacher's Note

When you buy items at a store with a total budget, you're solving simultaneous equations - finding the exact quantities of two products that fit your money constraint and shopping list.

Methods Of Solving Simultaneous Equations

To solve two simultaneous linear equations means, to find the values of variables used in the given equations.

Out of the different algebraic methods for solving simultaneous equations, we shall be discussing the following three methods only:

1. Method of elimination by substitution.

2. Method of elimination by equating coefficients.

3. Method of cross-multiplication.

Method Of Elimination By Substitution

Steps: 1. From any of the given two equations, find the value of one variable in terms of the other.

2. Substitute the value of the variable, obtained in step (1), in the other equation and solve it.

3. Substitute the value of the variable obtained in step (2), in the result of step (1) and get the value of the remaining unknown variable.

Solve the following system of equations using the method of elimination by substitution. \(x + y = 7\) and \(3x - 2y = 11\).

Step 1:

\(x + y = 7\) \(\Rightarrow\) \(y = 7 - x\)

Step 2:

\(3x - 2y = 11\) \(\Rightarrow\) \(3x - 2 (7 - x) = 11\)

\(\Rightarrow\) \(3x - 14 + 2x = 11\)

\(\Rightarrow\) \(5x = 25\) and \(x = 5\)

Step 3:

\(y = 7 - x\) \(\Rightarrow\) \(y = 7 - 5 = 2\)

Solution is: \(x = 5\) and \(y = 2\)

Alternative method:

Instead of finding the value of \(y\) in terms of \(x\); if we find the value of \(x\) in terms of \(y\) and proceed as above; the result will remain the same. For this:

Step 1:

\(x + y = 7\) \(\Rightarrow\) \(x = 7 - y\)

Step 2:

\(3x - 2y = 11\) \(\Rightarrow\) \(3(7 - y) - 2y = 11\)

\(\Rightarrow\) \(21 - 3y - 2y = 11\)

\(\Rightarrow\) \(-5y = -10\) and \(y = 2\)

Step 3:

\(x = 7 - y\) \(\Rightarrow\) \(x = 7 - 2 = 5\)

Solution is: \(x = 5\) and \(y = 2\)

Solve using elimination by substitution:

\(\frac{x+7}{5} - \frac{2x-y}{4} = 3y - 5\) and \(\frac{4x-3}{6} + \frac{5y-7}{2} = 18 - 5x\)

Solution:

\(\frac{x+7}{5} - \frac{2x-y}{4} = 3y - 5\) \(\Rightarrow\) \(\frac{4(x+7)-5(2x-y)}{20} = 3y - 5\)

i.e. \(4x + 28 - 10x + 5y = 60y - 100\)

\(\Rightarrow\) \(-6x - 55y = -128\)

i.e. \(6x = 128 - 55y\) and \(x = \frac{128-55y}{6}\)

\(\frac{4x-3}{6} + \frac{5y-7}{2} = 18 - 5x\) \(\Rightarrow\) \(\frac{4x-3+15y-21}{6} = 18 - 5x\)

i.e. \(4x + 15y - 24 = 108 - 30x\) \(\Rightarrow\) \(34x + 15y = 132\)

\(\Rightarrow\) \(34\left(\frac{128-55y}{6}\right) + 15y = 132\)

\(\Rightarrow\) \(34 \times 128 - 34 \times 55y + 90y = 132 \times 6\)

\(\Rightarrow\) \(4352 - 1870y + 90y = 792\)

\(\Rightarrow\) \(1780y = 3560\) i.e. \(y = \frac{3560}{1780} = 2\)

\(x = \frac{128-55y}{6} = \frac{128-55 \times 2}{6} = \frac{18}{6} = 3\)

Solution is: \(x = 3\) and \(y = 2\)

Teacher's Note

When solving real-world problems like mixing paint colors or adjusting recipe ingredients, substitution helps you express one requirement in terms of another to find the perfect balance.

Exercise 6 (A)

Solve the following pairs of linear (simultaneous) equations using method of elimination by substitution:

1. \(8x + 5y = 9\) and \(3x + 2y = 4\)

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

3. \(2x + 3y = 8\) and \(2x = 2 + 3y\)

4. \(0.2x + 0.1y = 25\) and \(2(x - 2) - 1.6y = 116\)

5. \(6x = 7y + 7\) and \(7y - x = 8\)

6. \(y = 4x - 7\) and \(16x - 5y = 25\)

7. \(2x + 7y = 39\) and \(3x + 5y = 31\)

8. \(1.5x + 0.1y = 6.2\) and \(3x - 0.4y = 11.2\)

9. \(2(x - 3) + 3(y - 5) = 0\) and \(5(x - 1) + 4(y - 4) = 0\)

10. \(\frac{2x+1}{7} + \frac{5y-3}{3} = 12\) and \(\frac{3x+2}{2} - \frac{4y+3}{9} = 13\)

Method Of Elimination By Equating Coefficients

Steps: 1. 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.

2. Add both the equations, as obtained in step 1, or subtract one equation from the other, so that the terms with equal numerical coefficient cancel mutually.

3. Solve the resulting equation to find the value of one of the unknowns.

4. Substitute this value in any of the two given equations and find the value of the other unknown.

Solve, using the method of elimination by equating coefficients: \(3x - 4y = 10\) and \(5x - 3y = 24\)

Solution:

\(3x - 4y = 10\) (i)

\(5x - 3y = 24\) (ii)

Step 1: Multiply equation (i) by 5 and equation (ii) by 3.

The resulting equations are:

\(15x - 20y = 50\)

\(15x - 9y = 72\)

Step 2: - + - [Subtracting]

\(-11y = -22\)

Step 3: \(y = 2\)

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