write one sòlution of the equation 2x_y+1=0. (OEQ)
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Answer:
Final Answer : h = 3/2 , k = 2
Steps:
1) DRAWING Graph :
We will find at least points on equation 2x-y +1=0
When x = 0, y = 1
When y = 0 , x = -0.5
Join these points to get the graph of equation :
2x- y + 1 =0,
That is, straight line.
2) If Graph passes through the point (h, 4) , then x = h, y = 5
So,
\begin{lgathered}2x - y + 1 = 0 \\ = > 2h - 4 + 1 = 0 \\ = > 2h - 3 = 0 \\ = > h = \frac{3}{2}\end{lgathered}
2x−y+1=0
=>2h−4+1=0
=>2h−3=0
=>h=
2
3
3) If graph passes through (1/2,k) ,then
x = 1/2 , y = k
So,
\begin{lgathered}2x - y+ 1 = 0 \\ = > 2( \frac{1}{2} ) - k + 1 = 0 \\ = >1 - k + 1 = 0 \\ = > k = 2\end{lgathered}
2x−y+1=0
=>2(
2
1
)−k+1=0
=>1−k+1=0
=>k=2
Hence,
\boxed {h = \frac{3}{2} \: and \: k = 2}
h=
2
3
andk=2
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