Example

Title: System of Linear Equations with no Solutions

Grade: 1300-a Lesson: S1-L8

Explanation: The best way to understand SAT-2 is by looking at some examples. Take turns and read each example for easy understanding.

Examples:

If the linear equation system kx - 2y = 10, 5x + 8y = 13, has no solution, then find the value of k.

Step 1a

The given equations are

kx - 2y = 10 eqution(1)

5x + 8y = 13 eqution(2)

The linear system is

\$a_1x + b_1 y = c_1\$ eqution(1)

\$a_2 x + b_2 y = c_2\$ eqution(2)

If there is no solution for both equation (1) and equation (2), we can express it as,

\$a_1/a_2 = b_1 / b_2 ne c_1 / c_2\$ equation(3)

Explanation: This condition indicates an inconsistent system of equations, meaning no standard solution exists. It is a valuable tool in determining the solvability of linear systems.

Step 1b

Compare linear equations

kx - 2y = 10 with equation (1) and

5x + 8y = 13 with equation (2)

So by comparing the coefficients, we can write

\$a_1 = k\$, \$a_2 = 5\$, \$b_1 = -2\$, \$b_2 = 8\$, \$ c_1 = 10\$

and \$c_2 = 13\$

Plug the values in equation (3)

\$ k / 5 = (-2) / 8 ne 10 / 13\$ …​. equation (4)

Explanation: Compare coefficients in linear equations, then substitute values in equation (3) then after simplification it becomes equation (4).

Step 1c

To find the value of k, take the first two parts of equation (4) and then solve it:

\$ k/5 = \ cancel (-2)^1 / \cancel(8)^4\$

\$ k = -1/ 4 times 5\$

\$k = -5 / 4\$

Explanation: An inconsistent system of equations means no standard solution exists. Compare coefficients and substitute values and simplify them to find the value of k. In this instance, k is equal to \$-5/4\$.


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