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3 years ago

Find all solutions for X and check. (different from the other question I asked earlier) thanks!

Mathematics

1 answer:

KiRa [710]3 years ago

4 0

$\displaystyle \\ 
17) \\ 
x(5x-10)=0 \\ 
x = 0 ~~\texttt{or}~~5x-10 = 0 \\ 
x_1 = \boxed{0} \\ \\ 
x_2 = \frac{10}{5} =\boxed{2} \\ 
\texttt{Check: } \\ 
x_1 = 0~~\Longrightarrow ~~ 0(5\cdot 0 - 10) = 0 \cdot (-10) = 0 \\ 
x_2 = 2~~\Longrightarrow ~~ 2(5\cdot 2 - 10) = 2(10 - 10) = 2 \cdot 0 = 0$

$\displaystyle 18) \\ 
x^2 + 7x-18 = 42 \\ 
x^2 + 7x-18 - 42 =0 \\ 
x^2 + 7x-60 =0 \\ 
.~~~~~7x = 12x - 5x \\ 
\underbrace{x^2 + 12x} -\underbrace{5x - 60} = 0 \\ \\ 
x(x+12) -5(x+12)=0 \\ 
(x+12)(x-5)=0 \\ 
x + 12 = 0 ~~\texttt{or}~~x-5 = 0 \\ \\ 
x_1 = \boxed{-12} \\ 
x_2 = \boxed{5} \\ \\ 
 \texttt{Check: } \\
x1 = -12 ~~\Longrightarrow~~ (-12)^2 + 7\cdot (-12)-18 = 144-84-18 = 42 \\ \\ 
x2 = 5 ~~\Longrightarrow~~ (5)^2 + 7\cdot (5)-18 = 25+35-18 = 42 \\ \\$

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Answer:

$x_{1} = \frac{176}{127} + \frac{71}{127}x_{4}\\\\ x_{2} = \frac{284}{127} + \frac{131}{254}x_{4}\\\\x_{3} = \frac{845}{127} + \frac{663}{254}x_{4}\\$

Step-by-step explanation:

As the given Augmented matrix is

$\left[\begin{array}{ccccc}9&-2&0&-4&:8\\0&7&-1&-1&:9\\8&12&-6&5&:-2\end{array}\right]$

Step 1 :

$r_{1}$ ↔ $r_{1} - r_{2}$

$\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&7&-1&-1&:9\\8&12&-6&5&:-2\end{array}\right]$

Step 2 :

$r_{3}$ ↔ $r_{3} - 8r_{1}$

$\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&7&-1&-1&:9\\0&124&-54&77&:-82\end{array}\right]$

Step 3 :

$r_{2}$ ↔ $\frac{r_{2}}{7}$

$\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&124&-54&77&:-82\end{array}\right]$

Step 4 :

$r_{1}$ ↔ $r_{1} + 14r_{2}$ , $r_{3}$ ↔ $r_{3} - 124r_{2}$

$\left[\begin{array}{ccccc}1&0&4&-11&:-8\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&0&- \frac{254}{7} &\frac{663}{7} &:-\frac{1690}{7} \end{array}\right]$

Step 5 :

$r_{3}$ ↔ $\frac{r_{3}. 7}{254}$

$\left[\begin{array}{ccccc}1&0&4&-11&:-8\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&0&1&-\frac{663}{254} &:-\frac{1690}{254} \end{array}\right]$

Step 6 :

$r_{1}$ ↔ $r_{1} - 4r_{3}$ , $r_{2}$ ↔ $r_{2} + \frac{1}{7} r_{3}$

$\left[\begin{array}{ccccc}1&0&0&-\frac{71}{127} &:\frac{176}{127} \\0&1&0&-\frac{131}{254} &:\frac{284}{127} \\0&0&1&-\frac{663}{254} &:\frac{845}{127} \end{array}\right]$

∴ we get

$x_{1} = \frac{176}{127} + \frac{71}{127}x_{4}\\\\ x_{2} = \frac{284}{127} + \frac{131}{254}x_{4}\\\\x_{3} = \frac{845}{127} + \frac{663}{254}x_{4}\\$

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3 years ago