Common Denominators 15. Explain how you can find a common denominator for and

Mathematics

1 answer:

VLD [36.1K]3 years ago

6 0

You can find a common denominator using the two denominators multiples.

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Which value, when placed in the box, would result in a system of equations with infinitely many solutions? y = -2x 4 6x 3y =

wariber [46]

The missing value is 12 in a system of equations with infinitely many solutions conditions.

It is given that in the system of equations there are two equations given:

$\rm y =n -2x+4\\\rm and \\\rm 6x+3y= ?$

It is required to find the missing value in the second equation.

<h3>What is a linear equation?</h3>

It is defined as the relation between two variables if we plot the graph of the linear equation we will get a straight line.

We have equations:

$\rm y = -2x+4 ....(1)\\ \\\rm 6x+3y= ? ....(2)$

Let's suppose the missing value is 'Z'

We know that the two pairs of equations have infinitely many solutions if and if they have the same coefficients of variables and the same constant on both sides.

From equation (1)

$\rm y = -2x+4 \\\\\rm 2x+y=4$ (multiply both the sides by 3)

$\rm 6x+3y=12$ ...(3)

By comparing the equation (2) and (3), we get

M = 12

Thus, the missing value is 12 in a system of equations with infinitely many solutions conditions.

Learn more about the linear equation.

brainly.com/question/11897796

8 0

2 years ago

1. Write an equation in slope-intercept form for a line passing through (-3,3) with a slope of 1.

PIT_PIT [208]

$\bf (\stackrel{x_1}{-3}~,~\stackrel{y_1}{3})~\hspace{10em} slope = m\implies 1 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-3=1[x-(-3)] \\\\\\ y-3=1(x+3)\implies y-3=x+3\implies y=x+6$

$\bf (\stackrel{x_1}{-4}~,~\stackrel{y_1}{2})\qquad (\stackrel{x_2}{1}~,~\stackrel{y_2}{12}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{12-2}{1-(-4)}\implies \cfrac{10}{1+4}\implies \cfrac{10}{5}\implies 2 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-2=2[x-(-4)]\implies y-2=2(x+4) \\\\\\ y-2=2x+8\implies y=2x+10$