Rivalry2

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Original 2026-01-01

Modified 2026-02-28

1 There are many methods for solving linear equations. Some of the methods are:

2 <ul><li>Graphical Method</li>

3 <li>Elimination Method</li>

4 <li>Substitution Method</li>

5 <li>Cross Multiplication Method</li>

6 <li>Matrix Method</li>

7 <li>Determinants Method (Cramer’s Rule) </li>

8 </ul>Graphical Method

9 For solving linear equations in a graphical method, we draw the equations as a line on a graph. The point where the two lines cross or intersect is the answer.

10 Example: \(y = 2x + 1\) \(y = x + 3\)

11 Plot both lines on the graph. Where they meet is the answer. If they meet at (2, 5), that means x = 2 and y = 5 is the solution.

12 Elimination Method

13 In this method, we add or subtract two equations to cancel out one variable, allowing us to find the value of the remaining variable.

14 Example: \(x + y = 10\) \(x - y = 4\) Add both equations, \((x + y) + (x - y) = 10 + 4\) \(2x = 14\) \(x = 7\) Now put x = 7 in the first equation: \(7 + y = 10\) \(y = 3\) So the final answer is \(x = 7, y = 3\)

15 Substitution Method

16 Substitution method means solving one equation to find one variable and then substituting it into the other.

17 Example: Solve the system: (1) \(y = x + 2\) (2) \(x + y = 10\)

18 Step 1:Use equation (1) to substitute for y in equation (2): \(x + (x + 2) = 10\)

19 Step 2:simplify the equation \(2x + 2 = 10\) \(2x = 8\) \(x = 4\)

20 Step 3:Substitute \(x = 4\) back into equation (1): \(y = 4 + 2 = 6\) Answer: \(x = 4, y = 6\)

21 Cross Multiplication Method

22 For solving linear equations using the<a>cross multiplication</a>method, we use<a>formulas</a>. The formula for solving two equations in the form of:

23 \(a_1x + b_1y = c_1\)

24 \(a_2x + b_2y = c_2\)

25 \(x = \frac{(b1c2 - b2c1)}{(a1b2 - a2b1)}, y = \frac{(c1a2 - c2a1)}{(a1b2 - a2b1)}\)

26 Matrix Method

27 A matrix method is a neat way to write equations in rows and columns. Let's see this using an example.

28 \(x + y = 6\) \(2x + 3y = 14\)

29 Write this in a matrix form as:

30 \( A = \begin{bmatrix} 1 & 1 \\ 2 & 3 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \end{bmatrix}, \quad B = \begin{bmatrix} 6 \\ 14 \end{bmatrix} \)

31 Now use the formula: \(AX = B\) \(X = A^{-1}B\)

32 Once we substitute the values and solve the equation, we get \(x = 4, y = 2\).

33 Determinants Method (Cramer’s Rule)

34 This method uses<a>determinants</a>, which are specific numerical values calculated from<a>square</a>matrices and help in solving systems<a>of equations</a>.

35 Example: \(2x + 3y = 12\) \(x - y = 1\)

36 We find three determinants: \( \Delta \) from coefficients \(\Delta_1\) replace the first column with answer numbers \(\Delta_2\) replace second column with answer numbers

37 \(x = \frac{\Delta_1}{\Delta}, y =\frac{\Delta_2}{\Delta}\)