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SIMULTANEOUS LINEAR EQUATIONS – SUBSTITUTION AND ELIMINATION METHODS

CONTENT

  1. Solving Simultaneous Linear Equations by Substitution
  2. Solving Simultaneous Linear Equations by Elimination
  3. Word Problems Leading to Simultaneous Linear Equations

 

Solving Simultaneous Linear Equations by Substitution

This is a method in which one variable is made the subject of the formula and substituted in the second equation. This will lead to a third equation with one variable. The equation is then solved for the value of the variable. This value can be substituted in any of the first two equations for the value of the second variable.

Example 1:

Solve \(2x + y = 4\)…………….i

\(3x  -y = -1\frac{1}{2}\)…………….ii

Using substitution method,

Solution:

\(2x + y = 4\) …………….i

\(3x -y = -\frac{3}{2}\)…………….ii

Considering equation 1 \((2x + y = 4)\),

\(y = 4 -2x\)

Now put \(y = 4 -2x\) into equation 2

Thus \(3x -(4 -2x) = -\frac{3}{2}\)…………….iii

\(3x -4 + 2x = -\frac{3}{2} \\ 5x -4 = -\frac{3}{2} \\ 5x = 4 -\frac{3}{2} = \frac{8 -3}{2} = \frac{5}{2} \\ 5x = \frac{5}{2} \\ 10x  = 5 \\ x = \frac{5}{10} = \frac{1}{2}\)

Substitute \(x = \frac{1}{2}\) into Eqn i

This gives \(2\Big(\frac{1}{2}\Big) + y = 4 \\ 1 + y = 4, \\ y = 4 -1 = 3\)

Thus \(x = \frac{1}{2}\) and \(y = 3\)

Check \(2\Big(\frac{1}{2}\Big) + 3 = 1 + 3 = 4 \) …………….Eqn 1

\(3\Big(\frac{1}{2}\Big) -3 = \frac{3}{2} -3 = \frac{3 -6}{2} = -\frac{3}{2}\) …………….Eqn 2

 

Example 2:

Solve \(3x -4y = -6\)  …………….Eqn 1

\(7x -2y = 8\) …………….Eqn 2

Using substitution method,

Solution:

\(3x -4y = -6\) …………….Eqn 1

\(7x -2y = 8\) …………….Eqn 2

From equation 1, make x the subject of the formula.

Lesson tags: JSS3 Mathematics, JSS3 Mathematics Evaluation Questions, JSS3 Mathematics Evaluation Questions Second Term, JSS3 Mathematics Objective Questions, JSS3 Mathematics Objective Questions Second Term, JSS3 Mathematics Second Term, Mathematics Lesson Notes, Mathematics Objective Questions
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