Use subtitution method to solve the problem.
First, change y to the value of x in order to find the exact value of x.
x + y = 4
y = 4 - x
Second, subtitute y with 4 - x from the first equation
y = -x² + 2x + 4
4 - x = -x² + 2x + 4
move all terms to the left side
x² - 2x - x + 4 - 4 = 0
x² - 3x = 0
x(x - 3) = 0
x = 0 or x = 3
Third, now we have 2 values of x. Find the value of y for each of x
For x = 0
y = 4 - x
y = 4 - 0
y = 4
For x = 3
y = 4 - x
y = 4 - 3
y = 1
The solution is (0,4) and (3,1). The answer is option b
If the discriminant of a quadratic equation, given by B^2-4AC is positive, it has 2 real solutions. If 0, it has 1 real solution and if negative it has 0 real solutions.
Answer:
- See attachment for table values
- y₁ = y₂ for x = 6
Step-by-step explanation:
In each case, put the x-value in the formula and do the arithmetic. If you're allowed, you can save some time and effort by realizing that the solution (x) will have to be an even number.
y₁ is an integer value for all integer values of x. y₂ is an integer value for even values of x only. y₁ and y₂ will both be integers (and possibly equal) only when x is even.
For example, for x = 6, we have
... y₁ = 3·6 - 8 = 18 -8 = 10
... y₂ = 0.5·6 +7 = 3 +7 = 10
That is, for x = 6, both columns of the table have the same number (10). That is, y₁ = y₂ for x = 6. The solution to the equation
... y₁ = y₂
is
... x = 6.