how do you solve that pair of simultaneous equation^2=4-y x=y+3?
38-13=??? 46+12=??? 56+13=??? 63-24=??? 67+25=??? 55-18=??? 85+56=??? 78-29=??? 92+37=???
To solve the pair of simultaneous equations:
x^2 = 4 - y ...(Equation 1)
x = y + 3 ...(Equation 2)
Let's proceed with the steps to solve these equations:
Step 1: Substitute the value of x from Equation 2 into Equation 1:
(y + 3)^2 = 4 - y
Step 2: Expand and simplify the equation:
y^2 + 6y + 9 = 4 - y
Step 3: Rearrange the equation to bring all terms to one side:
y^2 + 6y + y + 9 - 4 = 0
y^2 + 7y + 5 = 0
Step 4: Factorize or use the quadratic formula to solve the quadratic equation:
In this case, the equation does not factorize easily, so we will use the quadratic formula:
y = (-b ± √(b^2 - 4ac)) / (2a)
By comparing the equation y^2 + 7y + 5 = 0 with the standard quadratic equation ax^2 + bx + c = 0, we can identify:
a = 1, b = 7, c = 5
Substituting the values into the quadratic formula:
y = (-7 ± √(7^2 - 4*1*5)) / (2*1)
Simplifying further:
y = (-7 ± √(49 - 20)) / 2
y = (-7 ± √29) / 2
Hence, the solutions for y are:
y = (-7 + √29) / 2
y = (-7 - √29) / 2
Step 5: Substitute the values of y back into Equation 2 to find the corresponding values of x:
For y = (-7 + √29) / 2:
x = y + 3
x = ((-7 + √29) / 2) + 3
Simplifying further:
x = (-7 + √29 + 6) / 2
x = (-1 + √29) / 2
For y = (-7 - √29) / 2:
x = y + 3
x = ((-7 - √29) / 2) + 3
Simplifying further:
x = (-7 - √29 + 6) / 2
x = (-1 - √29) / 2
Therefore, the solutions to the pair of simultaneous equations are:
x = (-1 + √29) / 2, y = (-7 + √29) / 2
x = (-1 - √29) / 2, y = (-7 - √29) / 2