Let c be a constant. For what values of c can the simultaneous equations x – y = 2; cx + y = 3 have a solution (x, y) inside quadrant I
we need x>0 and y>0
eliminating y, we have
x-(3-cx) = 2
(c+1)x = 5
c+1 > 0
c > -1
eliminating x, we have
cx-cy = 2c
cx+y = 3
(1+c)y = 3-2c
y = (3-2c)/(c+1)
since c+1 > 0, 3-2c > 0, so c < 3/2
so, -1 < c < 3/2
To find the values of c for which the simultaneous equations have a solution (x, y) inside quadrant I, we need to determine the range of values for c that satisfy this condition.
First, let's rewrite the given system of equations:
Equation 1: x - y = 2
Equation 2: cx + y = 3
To find the solution (x, y) inside quadrant I, we need to consider the following conditions:
1. x > 0: The x-coordinate must be positive.
2. y > 0: The y-coordinate must be positive.
Let's solve the system of equations using these conditions and see how the values of c affect the solution.
Step 1: Solve Equation 1 for y.
x - y = 2
y = x - 2
Step 2: Substitute the value for y in Equation 2.
cx + y = 3
cx + (x - 2) = 3
(c + 1) x = 5
Step 3: Determine the range of values for c.
For a solution to exist in quadrant I, both x and y must be positive.
Condition 1: x > 0
Since x cannot be equal to zero, we have c + 1 ≠ 0.
Thus, c ≠ -1.
Condition 2: y > 0
Substitute the expression for y: y = x - 2.
x - 2 > 0
x > 2
Condition 3: Solve for c.
(c + 1) x = 5
We need both (c + 1) and x to be positive, and x > 2.
If (c + 1) > 0 (to satisfy y > 0), then c > -1.
Also, if x > 2, then (c + 1) x = 5 implies that (c + 1) > 0.
Therefore, c > -1.
Combining the conditions, we have:
c > -1.
Thus, for the simultaneous equations x - y = 2 and cx + y = 3 to have a solution (x, y) inside quadrant I, the value of c must be greater than -1.