4x-3y=9 8x+ky=19 For which value of k will the system of the equations have no solution?
I suppose you can try to solve it like a normal system, where k is some constant...
There is no solution when the lines are parallel. Parallel lines have the same slope. Let's find the slope of the first line by solving for y. \(4x - 3y = 9\) \(-3y = -4x + 9\) \(y = \dfrac{4}{3}x - 3\) \( \text{slope}_1 = m_1 = \dfrac{4}{3} \) Now let's find the slope of the second line. We solve it for y. \(8x+ky=19\) \(ky = -8x + 19\) \(y = -\dfrac{8}{k}x + \dfrac{19}{k} \) \( \text{slope}_2 = m_2 = \dfrac{4}{3} \) For the lines to be parallel, their slopes must be equal: \(m_1 = m_2\) \( \dfrac{4}{3} = -\dfrac{8}{k} \) \( k = -6 \)
thanks, i get it now! But how is the slope for the second equation 4/3 also. How did you solve i?
@mathstudent55
I wrote these two sentences above: "There is no solution when the lines are parallel. Parallel lines have the same slope." That is the key to solving this problem. The problem wants this system of equations to have no solution. Since parallel lines never intersect, if the two equations of the system of equations are parallel, then there will be no solution. We are given the equation of one of the lines. With that equation, we can find the slope of that line. That slope turned out to be 4/3. If we purposely make the slope of the second line also 4/3, then the lines will be parallel, and the system of equations will have no solution. When we solve the second equation for y, putting it in the y = mx + b form, the slope is m. The second equation in y = mx + b form turned out to be \(y = -\dfrac{8}{k}x + \dfrac{19}{k}\) The slope is what mulktiplies x, so the slope is \( -\dfrac{8}{k} \) The problem asks what value of k will ensure there is no solution. The value of k we are looking for is the value that will make the slope 4/3 (so the lines are parallel.) By solving the equation \( -\dfrac{8}{k} = \dfrac{4}{3} \) for k, we find the value of k that ensures the slope of the second line is 4/3.
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