Question

$x+y+z=25,\:5x+3y+2z=0,\:y-z=6$

solve it !!!!!!

Easy Difficult Question

Answer 1

Question :

☯ Solve   $\bf{x+y+z=25,\:5x+3y+2z=0,\:y-z=6}$

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✭ To Find :  

Value of x, y and z

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☞ Your Answer is  :

$\bf{x=-\dfrac{131}{5},\:z=\dfrac{113}{5},\:y=\dfrac{143}{5}}$

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Given :

✭ $\sf{x+y+z=25,\:5x+3y+2z=0,\:y-z=6}$

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Solution :

Given conditions :

${\begin{bmatrix}x+y+z=25\\ 5x+3y+2z=0\\ y-z=6\end{bmatrix}$

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$\mathrm{Isolate}\:y\:\mathrm{for}\:y-z=6$

$y-z=6$

$\mathrm{Add\:}z\mathrm{\:to\:both\:sides}$

$y-z+z=6+z$

$\mathrm{Simplify}$

$y=6+z$

$\mathrm{Subsititute\:}y=6+z$

$\begin{bmatrix}x+6+z+z=25\\ 5x+3\left(6+z\right)+2z=0\end{bmatrix}$

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$\mathrm{Simplify}$

$x+6+z+z=25\quad :\quad x+6+2z=25$

$5x+3\left(6+z\right)+2z=0\quad :\quad 5x+18+5z=0$

$= \begin{bmatrix}x+6+2z=25\\ 5x+18+5z=0\end{bmatrix}$

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$\mathrm{Isolate}\:x\:\mathrm{for}\:x+6+2z=25:\quad x=-2z+19$

$x+6+2z=25$

$\mathrm{Subtract\:}2z\mathrm{\:from\:both\:sides}$

$x+6+2z-2z=25-2z$

$\mathrm{Simplify}$

$x+6=25-2z$

$\mathrm{Subtract\:}6\mathrm{\:from\:both\:sides}$

$x+6-6=25-2z-6$

$\mathrm{Simplify}$

$x=-2z+19$

$\mathrm{Subsititute\:}x=-2z+19$

$\begin{bmatrix}5\left(-2z+19\right)+18+5z=0\end{bmatrix}$

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$\mathrm{Simplify}$

$5\left(-2z+19\right)+18+5z=0$

$\mathrm{Simplify}\:5\left(-2z+19\right)+18+5z:\quad -5z+113$

$\begin{bmatrix}-5z+113=0\end{bmatrix}$

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$\mathrm{Isolate}\:z\:\mathrm{for}\:-5z+113=0:\quad z=\frac{113}{5}$

$\mathrm{Subtract\:}113\mathrm{\:from\:both\:sides}$

$-5z+113-113=0-113$

$\mathrm{Simplify}$

$-5z=-113$

$z=\frac{113}{5}$

$\mathrm{For\:}x=-2z+19$

$\mathrm{Subsititute\:}x=-\frac{131}{5},\:z=\frac{113}{5}$

$y=6+\frac{113}{5}$

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$6+\frac{113}{5}=\frac{143}{5}$

$y=\frac{143}{5}$

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$\mathrm{The\:solutions\:to\:the\:system\:of\:equations\:are:}$

$\bf{x=-\dfrac{131}{5},\:z=\dfrac{113}{5},\:y=\dfrac{143}{5}}$

Answer 2

$\mathrm{Subtract\:}2z\mathrm{\:from\:both\:sides}$

$x+6+2z-2z=25-2z$

$\mathrm{Subtract\:}6\mathrm{\:from\:both\:sides}$

$x+6-6=25-2z-6$

$\mathrm{Subsititute\:}x=-2z+19$

$\begin{bmatrix}5\left(-2z+19\right)+18+5z=0\end{bmatrix}$

$5\left(-2z+19\right)+18+5z=0$

$\begin{bmatrix}-5z+113=0\end{bmatrix}$

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$\mathrm{Isolate}\:z\:\mathrm{for}\:-5z+113=0:\quad z=\frac{113}{5}$

[

$-5z+113-113=0-113$

$z=\frac{113}{5}$

$\mathrm{For\:}x=-2z+19$

$y=6+\frac{113}{5}$

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$6+\frac{113}{5}=\frac{143}{5}$

$y=\frac{143}{5}$