Lesson Example Discussion Quiz: Class Homework |
Example |
Title: System of Linear Equations with no Solutions |
Grade: 1400-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
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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) |
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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
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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) |
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Explanation: Compare coefficients in linear equations, then substitute values in equation (3) then after simplification it becomes equation (4). |
Step 1c
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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\$ |
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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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