how do you solve that pair of simultaneous equation^2=4-y x=y+3?
2 = 4 -y means y = 2.
Is that what you meant to type?
Once you know y, use
x=y+3
to find x.
NO what i meant was x^2=4-y
x=y+3
To solve the pair of simultaneous equations:
1) x^2 = 4 - y
2) x = y + 3
We can use substitution method or elimination method. Let's use substitution method.
Step 1: Solve Equation 2 for x
From Equation 2, we have x = y + 3.
Step 2: Substitute the value of x into Equation 1
Replace x in Equation 1 with y + 3:
(y + 3)^2 = 4 - y
Step 3: Expand and simplify Equation 1
By expanding (y + 3)^2, we get:
y^2 + 6y + 9 = 4 - y
Step 4: Rearrange and simplify Equation 1
Move all terms to one side of the equation:
y^2 + 6y + y + 9 - 4 = 0
y^2 + 7y + 5 = 0
Step 5: Solve Equation 1 for y
Apply quadratic formula:
y = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = 1, b = 7, and c = 5. Plugging in these values into the quadratic formula, we get:
y = (-7 ± √(7^2 - 4(1)(5))) / 2(1)
y = (-7 ± √(49 - 20)) / 2
y = (-7 ± √29) / 2
Therefore, the solutions for y are:
y = (-7 + √29) / 2
y = (-7 - √29) / 2
Step 6: Substitute the values of y into Equation 2 to find the corresponding values of x
Plugging the value of y into Equation 2, we get:
For y = (-7 + √29) / 2:
x = y + 3 = ((-7 + √29) / 2) + 3
For y = (-7 - √29) / 2:
x = y + 3 = ((-7 - √29) / 2) + 3
These are the solutions for the pair of simultaneous equations x^2 = 4 - y and x = y + 3.