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kupik [55]
3 years ago
5

2x-y=4 -2x+y=7 is there a solution

Mathematics
2 answers:
katovenus [111]3 years ago
7 0

\text{Given : 2x - y = 4 and x + 2y = 7}

\text{Solution : according to the question here we are provided with} \\ \text{two equations separately. So, let's name them as eq(1) and eq(2).} \\  \text{ The rest is followed as we continue to get to our goal}  \\  \\  \sf  {1}^{st}  \: equation  : \\   \quad\dashrightarrow \frak{2x - y = 4} \\  \\ \quad \dashrightarrow \frak{ \red{x =  \frac{y + 4}{2} }} \qquad  \sf ...eq(1) \\  \rm{ now \:  \: as \:per \: our \: plan \: we \: need \: to \: get \: eq(2) } \\  \\  \sf{ {2}^{nd}   \:equation} : \\   \quad\dashrightarrow \frak{ x + 2y = 7} \\  \\ \quad\dashrightarrow \frak{ \red{y =  \frac{7 - x}{2}}} \qquad \sf ...eq(2) \\  \\  \text{Now, finding x and y}  \\  \\  \underline{ \frak{Finding  \: x}}\\  \\  \quad \hookrightarrow \frak{x =  \frac{y  +  4}{2}}  \\  \\ \quad \hookrightarrow \frak{2x =  \frac{7 - x}{2} + 4} \\  \\ \quad \hookrightarrow \frak{2x =  \frac{7 - x + 8}{2}} \\  \\ \quad \hookrightarrow \frak{2x =  \frac{1 - x}{2}} \\  \\\quad \hookrightarrow \frak{4x =  1 - x}  \\  \\ \quad \hookrightarrow \frak{5x = 1}  \\  \\\quad \star \qquad \underline{ \boxed{ \green{ \frak{x =  \frac{1}{5}}}}} \\  \\  \underline{ \frak{Finding \:  y}} \\  \\  \quad \hookrightarrow \frak{y =  \frac{7 - x}{2}} \\  \\ \hookrightarrow \frak{2y =  7 -  \frac{1}{5} } \\  \\ \hookrightarrow \frak{2y =   \frac{35 - 1}{5} } \\  \\ \hookrightarrow \frak{2y =  \frac{34}{5}} \\  \\ \hookrightarrow \frak{y =   \frac{34}{5}  \times \frac{1}{2}} \\  \\ \hookrightarrow \frak{y =   \frac{17}{5} } \\  \\ \star \qquad \underline{ \boxed{ \frak{ \green{y = 3 \frac{2}{5}  }}}}

astra-53 [7]3 years ago
4 0
There is one solution which is: x= 3/2, y= 10. Or as in ordered pairs (3/2,10)
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Answer:

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Step-by-step explanation:

I assume you're asking how to make it into an equation form?

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