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nata0808 [166]
3 years ago
7

Which of the lines that appear in the graph would be parallel to a line with a slope of 3 and a y intercept at 0,3

Mathematics
1 answer:
Arlecino [84]3 years ago
3 0

Answer:

Line AB

Step-by-step explanation:

The line is on crosses thru the exact points so its right.

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We are given with an equation in <em>variable y</em> and we need to solve for <em>y</em> . So , now let's start !!!

We are given with ;

{:\implies \quad \sf \dfrac{33}{2}+\dfrac{3y}{5}=\dfrac{7y}{10}+15}

Take LCM on both sides :

{:\implies \quad \sf \dfrac{165+6y}{10}=\dfrac{7y+150}{10}}

<em>Multiplying</em> both sides by <em>10</em> ;

{:\implies \quad \sf \cancel{10}\times \dfrac{165+6y}{\cancel{10}}=\cancel{10}\times \dfrac{7y+150}{\cancel{10}}}

{:\implies \quad \sf 165+6y=7y+150}

Can be <em>further written</em> as ;

{:\implies \quad \sf 7y+150=165+6y}

Transposing <em>6y </em>to<em> LHS</em> and <em>150</em> to<em> RHS </em>

{:\implies \quad \sf 7y-6y=165-150}

{:\implies \quad \bf \therefore \quad \underline{\underline{y=15}}}

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\underline{+\left\{\begin{array}{ccc}8x+6y=28\\33x-6y=177\end{array}\right}\ \ \ \ |\text{add both sides of the equations}\\.\ \ \ \ \ \ \ 41x=205\ \ \ \ |:41\\.\ \ \ \ \ \ \ x=5\\\\\text{substitute the value of x to the first equation}\\\\4(5)+3y=14\\20+3y=14\ \ \ \ |-20\\3y=-6\ \ \ \ \ |:3\\y=-2\\\\\text{substitute the value of x to the equation:}\ z=12-3x\\\\z=12-3(5)=12-15=-3\\\\x=5,\ y=-2,\ z=-3\to(5,\ -2,\ -3)

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