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Example1

Title: System of Linear Equations with no Solutions

Grade Lesson s5-l8

Explanation: The best way to understand SAT-4 is by looking at some examples. Take turns and read each example for easy understanding.

Examples

Topics → Definition Example1

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

Step: 1

The given equation is kx - 2y = 10 and 5x + 8y = 13

Here we consider two linear equations formula

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

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

If equation (1) and equation(2) have no solution then we can write,

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

Explanation:

Here we consider two linear equations and simplify in into equation (3).

Step: 2

First, we will compare 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\$

Substitute the values in the formula.: \$ k / 5 = (-2) / 8 ne (10) /(13)\$

Explanation:

Here we substitute the vales in the formula.

Step: 3

Now taking the first two parts of the above equation, we get: \$ k/5 = (\cancel (-2)^1)/(\cancel(8)^4)\$

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

\$k = -5/4\$

Explanation:

Here simplify the equation, we get the k value is \$-5/4\$.