For which t-values does the system have none, one or infinite solutions? 4(t+4)x+21y=56 ;color(white)("dd")12x+3ty=24

I am unsure on how to solve this one with for example setting up a matrix, could someone help me on this one? $4 \left(t + 4\right) x + 21 y = 56$ $12 x + 3 t y = 24$

May 3, 2018

when the system has no solutions, $t = - 7$
when the system has infinite solutions, $7 = 3$
when the system has one solution, $t > - 7 < t < 3 < t$

Explanation:

Before we start, know that having either one or infinite solutions equals having the same gradient from both equations, and that having one solution equals having different gradients from both equations, with each event being mutually exclusive.

$4 \left(t + 4\right) x + 21 y = 56$ $\rightarrow$

$12 x + 3 t y = 24$ $\rightarrow$

Simplify,

$4 t x + 16 x + 21 y = 56$ $\rightarrow$

$12 x + 3 t y = 24$ $\rightarrow$

Method 1: Slope

Simplify the equations into slope-intercept form,

$y = \left(\frac{- 4 t - 16}{21}\right) x + \frac{56}{21}$ $\rightarrow$

$y = \left(- \frac{4}{t}\right) x + \frac{8}{t}$ $\rightarrow$

Find the gradient of both equations using the slope-intercept form,

Gradient of ①$= \frac{- 4 t - 16}{21}$

Gradient of ②$= - \frac{4}{t}$

To find having none or infinite solutions, the gradients of both equations are equal,

Gradient of ①$=$Gradient of ②

$\frac{- 4 t - 16}{21} = - \frac{4}{t}$

$\frac{4 t + 16}{21} = \frac{4}{t}$

$4 {t}^{2} + 16 t - 84 = 0$

${t}^{2} + 4 t - 21 = 0$

$\left(t - 3\right) \left(t + 7\right) = 0$

$t = 3 \mathmr{and} - 7$

Now we know that when $t = 3 \mathmr{and} - 7$, there is either no solutions or infinite solutions, but we don't know which is which.

Substitute $t = 3$ into ①$=$

$\left(- \frac{4}{3}\right) x + \frac{8}{3} = \left(\frac{- 4 \left(3\right) - 16}{21}\right) x + \frac{56}{21}$

$- \frac{4}{3} x + \frac{8}{3} = - \frac{28}{21} x + \frac{56}{21}$

$0 x = 0$

$x \in \setminus m a t h \boldsymbol{R}$

Substitute $t = - 7$ into ①$=$

$\left(- \frac{4}{- 7}\right) x + \frac{8}{- 7} = \left(\frac{- 4 \left(- 7\right) - 16}{21}\right) x + \frac{56}{21}$

$\frac{4}{7} x - \frac{8}{7} = \frac{12}{21} x + \frac{56}{21}$

$0 x = \frac{80}{21}$

$x = \text{N.A.}$

When the system has no solutions, $t = - 7$ and when the system has infinite solutions, $7 = 3$,

Find $t$, when the system has one solution,

$t > - 7 < t < 3 < t$

This is because this event is mutually exclusive as the other two events, meaning $t$ can be anything except $- 7$ and $3$.

Therefore, when the system has no solutions, $t = - 7$, when the system has infinite solutions, $7 = 3$, and when the system has one solution, $t > - 7 < t < 3 < t$.

Method 2: Matrix

${\underbrace{\left(\begin{matrix}4 t + 16 & 21 \\ 12 & 3 t\end{matrix}\right)}}_{A} {\underbrace{\left(\begin{matrix}x \\ y\end{matrix}\right)}}_{X} = {\underbrace{\left(\begin{matrix}56 \\ 24\end{matrix}\right)}}_{B}$

$A X = B$

To find $t$, you have to find $X$,

$X = {A}^{- 1} B$

Find ${A}^{- 1}$,

${A}^{- 1} = \left(\begin{matrix}\frac{t}{4 {t}^{2} - 5} & - \frac{7}{4 {t}^{2} - 5} \\ - \frac{1}{{t}^{2} - 17} & \frac{t + 4}{3 {t}^{2} - 9}\end{matrix}\right)$ ( using an inverse matrix calculator )

Find $X$

$X = {A}^{- 1} B$
$\textcolor{w h i t e}{X} = \left(\begin{matrix}\frac{t}{4 {t}^{2} - 5} & - \frac{7}{4 {t}^{2} - 5} \\ - \frac{1}{{t}^{2} - 17} & \frac{t + 4}{3 {t}^{2} - 9}\end{matrix}\right) \left(\begin{matrix}56 \\ 24\end{matrix}\right)$
$\textcolor{w h i t e}{X} = \left(\begin{matrix}\frac{14}{t + 7} \\ \frac{8}{t + 7}\end{matrix}\right)$

May 3, 2018

The system has no solutions for $t = - 7 \mathmr{and} t = 3$
The system has a unique solution for all other values of t.

Explanation:

The system has none or infinite solutions when the determinant of the coefficients is equal to 0:

| (4(t+4),21), (12,3t) | = 0

$\left(3 t\right) \left(4 \left(t + 4\right)\right) - \left(12\right) \left(21\right) = 0$

$12 {t}^{2} + 48 t - 252 = 0$

${t}^{2} + 4 t - 21 = 0$

$\left(t + 7\right) \left(t - 3\right) = 0$

$t = - 7 \mathmr{and} t = 3$

When $t = - 7$ the following is the augmented matrix:

[ (-12,21,|,56), (12,-21,|,24) ]

Please observe that if we multiply Row 2 by -1 we obtain parallel lines that do not overlay, therefore, there is no solution:

[ (-12,21,|,56), (-12,21,|,-24) ]

When $t = 3$ the following is the augmented matrix:

[ (21,21,|,56), (12,12,|,24) ]

Please observe that if we divide Row 1 by 21 and Row 2 by 12 we, again, obtain parallel lines that do not overlay, therefore, there is no solution:

[ (1,1,|,56/21), (1,1,|,2) ]

The system has a unique solution for all other values of t.