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# Homework Help ASAP (Unsolved)

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1) For a certain value of $$k$$, the system \begin{align*} x + y + 3z &= 10, \\ -4x + 2y + 5z &= 7, \\ kx + z &= 3 \end{align*}

has no solutions. What is this value of $$k$$?

2) Three points are on the same line if the slope of the line through the first two points is the same as the slope of the line through the second two points. For what value of $$c$$ are the three points$$(2,4), (6,3),$$ and $$(-5,c)$$ on the same line?

Jul 31, 2018
edited by Guest  Jul 31, 2018
edited by Guest  Jul 31, 2018
edited by Guest  Jul 31, 2018

#1
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1)

Solve the following system for x, y and z:
{x + y + 3 z = 10 | (equation 1)
-4 x + 2 y + 5 z = 7 | (equation 2)
k x + z = 3 | (equation 3)

Swap equation 1 with equation 2:
{-(4 x) + 2 y + 5 z = 7 | (equation 1)
x + y + 3 z = 10 | (equation 2)
k x + 0 y+z = 3 | (equation 3)

Add 1/4 × (equation 1) to equation 2:
{-(4 x) + 2 y + 5 z = 7 | (equation 1)
0 x+(3 y)/2 + (17 z)/4 = 47/4 | (equation 2)
k x + 0 y+z = 3 | (equation 3)

Multiply equation 2 by 4:
{-(4 x) + 2 y + 5 z = 7 | (equation 1)
0 x+6 y + 17 z = 47 | (equation 2)
k x + 0 y+z = 3 | (equation 3)

Add k/4 × (equation 1) to equation 3:
{-(4 x) + 2 y + 5 z = 7 | (equation 1)
0 x+6 y + 17 z = 47 | (equation 2)
0 x+(k y)/2 + ((5 k)/4 + 1) z = (7 k)/4 + 3 | (equation 3)

Multiply equation 3 by 4:
{-(4 x) + 2 y + 5 z = 7 | (equation 1)
0 x+6 y + 17 z = 47 | (equation 2)
0 x+2 k y + (5 k + 4) z = 7 k + 12 | (equation 3)

Subtract k/3 × (equation 2) from equation 3:
{-(4 x) + 2 y + 5 z = 7 | (equation 1)
0 x+6 y + 17 z = 47 | (equation 2)
0 x+0 y+(4 - (2 k)/3) z = 12 - (26 k)/3 | (equation 3)

Multiply equation 3 by 3/2:
{-(4 x) + 2 y + 5 z = 7 | (equation 1)
0 x+6 y + 17 z = 47 | (equation 2)
0 x+0 y+(6 - k) z = 18 - 13 k | (equation 3)

Divide equation 3 by 6 - k:
{-(4 x) + 2 y + 5 z = 7 | (equation 1)
0 x+6 y + 17 z = 47 | (equation 2)
0 x+0 y+z = (13 k - 18)/(k - 6) | (equation 3)

Subtract 17 × (equation 3) from equation 2:
{-(4 x) + 2 y + 5 z = 7 | (equation 1)
0 x+6 y+0 z = (-6 (29 k - 4))/(k - 6) | (equation 2)
0 x+0 y+z = (13 k - 18)/(k - 6) | (equation 3)

Divide equation 2 by 6:
{-(4 x) + 2 y + 5 z = 7 | (equation 1)
0 x+y+0 z = (4 - 29 k)/(k - 6) | (equation 2)
0 x+0 y+z = (13 k - 18)/(k - 6) | (equation 3)

Subtract 2 × (equation 2) from equation 1:
{-(4 x) + 0 y+5 z = (65 k - 50)/(k - 6) | (equation 1)
0 x+y+0 z = (4 - 29 k)/(k - 6) | (equation 2)
0 x+0 y+z = (13 k - 18)/(k - 6) | (equation 3)

Subtract 5 × (equation 3) from equation 1:
{-(4 x)+0 y+0 z = 40/(k - 6) | (equation 1)
0 x+y+0 z = (4 - 29 k)/(k - 6) | (equation 2)
0 x+0 y+z = (13 k - 18)/(k - 6) | (equation 3)

Divide equation 1 by -4:
{x+0 y+0 z = (-10)/(k - 6) | (equation 1)
0 x+y+0 z = (4 - 29 k)/(k - 6) | (equation 2)
0 x+0 y+z = (13 k - 18)/(k - 6) | (equation 3)

x = -10/(k - 6)
y = (4 - 29 k)/(k - 6)
z = (13 k - 18)/(k - 6)     k=6, for which there is NO solution.

Jul 31, 2018
#2
+21244
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1)

For a certain value of , the system

\begin{align*} x + y + 3z &= 10, \\ -4x + 2y + 5z &= 7, \\ kx + z &= 3 \end{align*}

\begin{align*} x + y + 3z &= 10, \\ -4x + 2y + 5z &= 7, \\ kx + z &= 3 \end{align*}

has no solutions. What is this value of $$k$$ ?

$$\text{Let \det(A) = \begin{vmatrix} 1 & 1 & 3 \\ -4& 2& 5 & \\ k& 0& 1 \end{vmatrix} }\\ \text{Let b = \begin{pmatrix} 10 \\ 7 \\ 3 \end{pmatrix} }\\ \text{If \det(A)=0, then the linear system Ax=b, b\ne0, has no solution. }$$

$$\begin{array}{|rcll|} \hline \det(A) = \begin{vmatrix} 1 & 1 & 3 \\ -4& 2& 5 & \\ k& 0& 1 \end{vmatrix} &=& 0 \\\\ \begin{vmatrix} 1 & 1 & 3 \\ -4& 2& 5 & \\ k& 0& 1 \end{vmatrix} &=& 0 \\\\ 1\cdot \begin{pmatrix} 2 & 5 \\ 0& 1 \end{pmatrix} -1\cdot \begin{pmatrix} -4 & 5 \\ k& 1 \end{pmatrix} +3\cdot \begin{pmatrix} -4 & 2 \\ k& 0 \end{pmatrix} &=& 0 \\\\ 2-(-4-k)+3(-2k) &=& 0 \\ 2+4+5k-6k &=& 0 \\ 2+4-k &=& 0 \\ k &=& 2+4 \\ \mathbf{k} &\mathbf{=}& \mathbf{6} \\ \hline \end{array}$$

Aug 1, 2018
#3
+21244
+1

2)
Three points are on the same line if the slope of the line through the first two points is the same as
the slope of the line through the second two points.
For what value of $$c$$ are the three points
$$(2,4), (6,3),$$
and
$$(-5,c)$$
on the same line?

$$\text{Let P_1=(2,4) }\\ \text{Let P_2=(6,3) }\\ \text{Let P_3=(-5,c) }$$

$$\begin{array}{|rcll|} \hline \dfrac{y_2-y_1}{x_2-x_1} &=& \dfrac{y_3-y_2}{x_3-x_2} \\\\ \dfrac{3-4}{6-2} &=& \dfrac{c-3}{-5-6} \\\ -\dfrac{1}{4} &=& -\dfrac{c-3}{11} \\\\ \dfrac{1}{4} &=& \dfrac{c-3}{11} \\\\ c-3 &=& \dfrac{11}{4} \\\\ c &=& 3+ \dfrac{11}{4} \\\\ c &=& \dfrac{12+11}{4} \\\\ c &=& \dfrac{23}{4} \\\\ \mathbf{c} &\mathbf{=}& \mathbf{5.75} \\ \hline \end{array}$$

Aug 1, 2018