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Lady bird [3.3K]
2 years ago
6

Find a and b if the point p(6,0) and Q(3,2) lie on the graph of ax+ by=12

Mathematics
1 answer:
tiny-mole [99]2 years ago
5 0

to get the equation of any straight line we only need two points off of it, hmmm let's use P and Q here and then let's set the equation in standard form, that is

standard form for a linear equation means

• all coefficients must be integers, no fractions

• only the constant on the right-hand-side

• all variables on the left-hand-side, sorted

• "x" must not have a negative coefficient

(\stackrel{x_1}{6}~,~\stackrel{y_1}{0})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{2}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{2}-\stackrel{y1}{0}}}{\underset{run} {\underset{x_2}{3}-\underset{x_1}{6}}}\implies \cfrac{2}{-3}\implies -\cfrac{2}{3} \\\\\\ \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-\stackrel{y_1}{0}=\stackrel{m}{-\cfrac{2}{3}}(x-\stackrel{x_1}{6})

\stackrel{\textit{multiplying both sides by }\stackrel{LCD}{3}}{3(y-0)~~ = ~~3\left( -\cfrac{2}{3}(x-6) \right)}\implies 3y=-2(x-6) \\\\\\ 3y=-2x+12 \implies \stackrel{a}{2} x+\stackrel{b}{3} y=12

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Which statements are true the ordered pair (1, 2) and the system of equations?
Delvig [45]
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
Below are the answers:
<span>When (1, 2) is substituted into the first equation, the equation is false.
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The ordered pair (1, 2) is not a solution to the system of linear equations.

The ordered pair (1, 2) is a solution to the system of linear equations.</span>
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3 years ago
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6 0
3 years ago
0.5x -2 = 1.5x + 0.15
frutty [35]

Answer: x=−2.15

Step-by-step explanation : Alright Let's solve your equation step-by-step.

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Step 1: Subtract 1.5x from both sides.

0.5x−2−1.5x=1.5x+0.15−1.5x

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Step 2: Add 2 to both sides.

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Step 3: Divide both sides by -1.

−x

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x=−2.15<em>        hope this helped!- Alex</em><u> </u>

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