{:[-6.4 x=4y+2.1],[ky+3.2 x=5.8]:} For what value of k does the system of linear equations in the variables x and y have no solutions?

Question

{:[-6.4 x=4y+2.1],[ky+3.2 x=5.8]:} For what value of  k does the system of linear equations in the variables  x and  y have no solutions?

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Conditions for no solutions: Understand the conditions for no solutions in a system of linear equations. A system of linear equations has no solutions when the lines represented by the equations are parallel. This means that the coefficients of x and y in both equations must be proportional, but the constants on the right side of the equations must not be proportional.Given system of equations: Write down the given system of equations. The system of equations is: -6.4x = 4y + 2.1 ky + 3.2x = 5.8Rearranging the equations: Rearrange the equations to have the same structure. We want to express both equations in the form of y = mx + b, where m is the slope and b is the y-intercept. For the first equation: 4y = -6.4x - 2.1 y = (-6.4/4)x - (2.1/4) y = -1.6x - 0.525 For the second equation: ky = -3.2x + 5.8 y = (-3.2/k)x + (5.8/k)Comparing the slopes: Compare the slopes of the two equations. For the lines to be parallel, the slopes of the two equations must be equal. The slope of the first equation is -1.6. The slope of the second equation is -3.2/k. So, we set -1.6 equal to -3.2/k: -1.6 = -3.2/kSolving for k: Solve for k. Multiply both sides by k to get rid of the fraction: k(-1.6) = -3.2 -1.6k = -3.2 Now, divide both sides by -1.6 to solve for k: k = -3.2 / -1.6 k = 2Checking if constants are proportional: Check if the constants are proportional. Now that we have found a value for k, we need to check if the constants on the right side of the equations are proportional. If they are not, then the lines are parallel and there are no solutions. For the first equation, the constant is -0.525. For the second equation, the constant is 5.8/k, which is 5.8/2 = 2.9. Since -0.525 is not proportional to 2.9, the lines are parallel and there are no solutions.

$\begin{aligned} -6.4 x & =4 y+2.1 \\ k y+3.2 x & =5.8 \end{aligned}$ $\newline$ For what value of $k$ does the system of linear equations in the variables $x$ and $y$ have no solutions?

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−6.4xamp;=4y+2.1ky+3.2xamp;=5.8 \begin{aligned} -6.4 x &amp; =4 y+2.1 \\ k y+3.2 x &amp; =5.8 \end{aligned} −6.4xky+3.2x​amp;=4y+2.1amp;=5.8​\newlineFor what value of k k k does the system of linear equations in the variables x x x and y y y have no solutions?

$\begin{aligned} -6.4 x & =4 y+2.1 \\ k y+3.2 x & =5.8 \end{aligned}$ $\newline$ For what value of $k$ does the system of linear equations in the variables $x$ and $y$ have no solutions?