Question

3x+2y +25 =0and 2x+y+10=0 by unique method​

Answer 1

Answer:

$3x+2y+25=0,\:2x+y+10=0\quad :\quad x=5,\:y=-20$

Step-by-step explanation:

Let's Solve this Question using Gaussian elimination:

$\begin{bmatrix}3x+2y+25=0\\ 2x+y+10=0\end{bmatrix}$

$\blue{\mathrm{Isolate\:solutions}}$

$\gray{\mathrm{Isolate\:solution}:\quad 3x+2y+25=0}$

$3x+2y=-25$

$\gray{\mathrm{Isolate\:solution}:\quad 2x+y+10=0}$

$2x+y=-10$

$\begin{bmatrix}3x+2y=-25\\ 2x+y=-10\end{bmatrix}$

$\gray{\mathrm{Write\:a\:matrix\:with\:the\:coefficients\:and\:solutions}}$

$\begin{bmatrix}3&2&\gray{|}&-25\\ 2&1&\gray{|}&-10\end{bmatrix}$

$\blue{\mathrm{Reduce\:matrix\:to\:reduced\:row\:echelon\:form}\:\begin{pmatrix}1\:&\:\cdots \:&\:b\:\\ 0\:&\ddots \:&\:\vdots \\ 0\:&\:0\:&\:1\end{pmatrix}}$

$\begin{pmatrix}3&2&-25\\ 2&1&-10\end{pmatrix}$

$\green{\mathrm{Reduce\:matrix\:to\:row\:echelon\:form}\:\begin{pmatrix}a&\cdots &b\\ 0&\ddots &\vdots \\ 0&0&c\end{pmatrix}}$

$\gray{\mathrm{Cancel\:leading\:coefficient\:in\:row\:}\:R_2\:\mathrm{\:by\:performing}\:R_2\:\leftarrow \:R_2-\frac{2}{3}\cdot \:R_1}$

$\displaystyle=\begin{pmatrix}3&2&-25\\ 0&-\displaystyle\frac{1}{3}&\displaystyle\frac{20}{3}\end{pmatrix}$

$\blue{\mathrm{Reduce\:matrix\:to\:reduced\:row\:echelon\:form}\:\begin{pmatrix}1\:&\:\cdots \:&\:b\:\\ 0\:&\ddots \:&\:\vdots \\ 0\:&\:0\:&\:1\end{pmatrix}}$

$\gray{\mathrm{Multiply\:matrix\:row\:by\:constant:}\:R_2\:\leftarrow \:-3\cdot \:R_2}$

$=\begin{pmatrix}3&2&-25\\ 0&1&-20\end{pmatrix}$

$\gray{\mathrm{Cancel\:leading\:coefficient\:in\:row\:}\:R_1\:\mathrm{\:by\:performing}\:R_1\:\leftarrow \:R_1-2\cdot \:R_2}$

$=\begin{pmatrix}3&0&15\\ 0&1&-20\end{pmatrix}$

$\displaystyle\gray{\mathrm{Multiply\:matrix\:row\:by\:constant:}\:R_1\:\leftarrow \frac{1}{3}\cdot \:R_1}$

$=\begin{pmatrix}1&0&5\\ 0&1&-20\end{pmatrix}$

$\begin{bmatrix}1&0&\gray{|}&5\\ 0&1&\gray{|}&-20\end{bmatrix}$

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

$x=5,\:y=-20$

There are many methods, by which you can solve this question. Let's see another way of solving this question:

That is using the Cramer's rule:

$\begin{bmatrix}3x+2y+25=0\\ 2x+y+10=0\end{bmatrix}$

$\gray{\mathrm{Isolate\:solutions}}$

$\begin{bmatrix}3x+2y=-25\\ 2x+y=-10\end{bmatrix}$

$\gray{\mathrm{Matrix\:of\:Coefficients}}$

$M=\begin{pmatrix}3&2\\ 2&1\end{pmatrix}$

$\gray{\mathrm{Answers\:column:}}$

$\begin{pmatrix}-25\\ -10\end{pmatrix}$

$\gray{\mathrm{Replace\:the\:}x\mathrm{-column\:values\:with\:the\:answer-column\:values}}$

$M_x=\begin{pmatrix}-25&2\\ -10&1\end{pmatrix}$

$\gray{\mathrm{Replace\:the\:}y\mathrm{-column\:values\:with\:the\:answer-column\:values}}$

$M_y=\begin{pmatrix}3&-25\\ 2&-10\end{pmatrix}$

$\blue{D=-1}$

$\blue{D_x=-5}$

$\blue{D_y=20}$

$\gray{\mathrm{Solve\:by\:using\:Cramer\:Rule}}$

$\displaystyle\gray{x=\frac{D_x}{D},\:y=\frac{D_y}{D},\:z=\frac{D_z}{D}}$

$\gray{D\:\mathrm{denotes\:the\:determinant}}$

$\displaystyle x=\frac{D_x}{D}=\frac{-5}{-1}$

$\gray{\mathrm{Simplify}}$

$x=5$

$\displaystyle y=\frac{D_y}{D}=\frac{20}{-1}$

$\gray{\mathrm{Simplify}}$

$y=-20$

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

$x=5,\:y=-20$