In which graph is y a linear function of x will mark brainest ty so much

Mathematics

1 answer:

pentagon [3]3 years ago

6 0

a? I'm unsure abt this one

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Can a linear equation and a linear inequality be solved in the same way

frez [133]

You can solve them in the same way. you can add any positive or negative number to both the sides of an inequality.Also you can multiply or divide both sides of an inequality by any positive number hope this helps. :)

8 0

3 years ago

<img src="https://tex.z-dn.net/?f=%5Cleft.%20%5Cbegin%7Bcases%7D%20%7B%208%20x%20%2B%202%20y%20%3D%2046%20%7D%20%5C%5C%20%7B%207

Brut [27]

$\huge \boxed{\mathfrak{Question} \downarrow}$

$\left. \begin{cases} { 8 x + 2 y = 46 } \\ { 7 x + 3 y = 47 } \end{cases} \right.$

$\large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}$

$\left. \begin{cases} { 8 x + 2 y = 46 } \\ { 7 x + 3 y = 47 } \end{cases} \right.$

First, write both the equations in its standard form.

$8x+2y=46\\ 7x+3y=47$

Now, write the equations in form of matrix.

$\left(\begin{matrix}8&2\\7&3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}46\\47\end{matrix}\right)$

Then, multiply the equation towards the left by using the inverse of matrix $\left(\begin{matrix}8&2\\7&3\end{matrix}\right)$

$\sf \: inverse(\left(\begin{matrix}8&2\\7&3\end{matrix}\right))\left(\begin{matrix}8&2\\7&3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}8&2\\7&3\end{matrix}\right))\left(\begin{matrix}46\\47\end{matrix}\right)$

The product of the matrix & its inverse will be the identity matrix.

$\sf\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}8&2\\7&3\end{matrix}\right))\left(\begin{matrix}46\\47\end{matrix}\right)$

Now, multiply the matrices that lie on the left-hand side of the equal sign.

$\sf\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}8&2\\7&3\end{matrix}\right))\left(\begin{matrix}46\\47\end{matrix}\right)$

For the 2 × 2 matrix $\left(\begin{matrix}a&b\\c&d\end{matrix}\right)$ , the inverse matrix is ⇨ $\left(\begin{matrix}\frac{d}{ad-bc}&\frac{-b}{ad-bc}\\\frac{-c}{ad-bc}&\frac{a}{ad-bc}\end{matrix}\right)$ .

$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{8\times 3-2\times 7}&-\frac{2}{8\times 3-2\times 7}\\-\frac{7}{8\times 3-2\times 7}&\frac{8}{8\times 3-2\times 7}\end{matrix}\right)\left(\begin{matrix}46\\47\end{matrix}\right)$

Do the calculations.

$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{10}&-\frac{1}{5}\\-\frac{7}{10}&\frac{4}{5}\end{matrix}\right)\left(\begin{matrix}46\\47\end{matrix}\right)$

Multiply the matrices.

$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{10}\times 46-\frac{1}{5}\times 47\\-\frac{7}{10}\times 46+\frac{4}{5}\times 47\end{matrix}\right)$