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Question

Find a range of values of c such that the lines x-y=2 and cx+y=3 intersect in the first quadrant.


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Solution

Step 1: Define the problem

Given two lines x-y=2 and cx+y=3.

Since the intersection happens in the first quadrant, x>0 and y>0.

From the first equation,

x-y=2⇒x=y+2

Step 2: Compute range of c for y>0

Substituting x=y+2 in cx+y=3,

c(y+2)+y=3⇒cy+2c+y=3⇒y(c+1)=3-2c⇒y=3-2cc+1

Therefore, for y>0, c∈-1,32

Step 3: Compute range of c for x>0

Substituting y=3-2cc+1 in x-y=2

x-3-2cc+1=2⇒x=5c+1

Therefore, for x>0, c∈-1,∞

Step 4: Compute range of values for c

c∈-1,32∩-1,∞⇒c∈-1,32

Therefore, the range of values for c such that the two lines intersect in first quadrant is, -1<c<32.


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