Question

Determine all values of h and k for which the system S 1 -3x - 3y = h -4x + ky = 10 has no solution. k= ht

Answer 1

Answer:

$h\neq 7.5$

k = -4

Step-by-step explanation:

Given system of equations are,

-3x-3y = h

-4x + ky = 10

We have to find the values of h and k such that system of equations has no solution.

The standard form of system of equation in two variables can be given by,

$a_1x+b_1y+c_1=0$

$a_2x+b_2y+c_2=0$

And condition for the system of equations has no solution is given by,

$\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}\neq \dfrac{c_1}{c_2}$

So, by comparing the standard form of equations with given equations, the condition such that system has no solution can be written as,

$\dfrac{-3}{-4}=\dfrac{-3}{k}\neq \dfrac{h}{10}$

$=>\dfrac{-3}{-4}=\dfrac{-3}{k}$

=> k = -4

and $\dfrac{-3}{-4}\neq \dfrac{h}{10}$

    $=>\ \dfrac{-3\times 10}{-4}\neq h$

    $=>\ h\neq \dfrac{30}{4}$

    $=>\ h\neq 7.5$

So, the value of h and k for above given system of equations is

$h\neq 7.5$ and k = -4.