The solutions to the given system of equations is (0, -6) and (1, -5)
<h3>Simultaneous equations</h3>
From the question, we are to determine the solutions to the given system of equations
The equations are
x − y = 6 --------- (1)
y = x² −6 ---------- (2)
From equation (1)
x - y = 6
∴ x = 6 + y ------- (3)
Substitute into equation (2)
y = x² −6
y = (6+y)² −6
y = (6+y)(6+y) -6
y = 36 + 6y + 6y +y² -6
y = 36 + 12y + y² - 6
Simplifying
y² + 12y - y + 30 = 0
y² + 11y + 30 = 0
Solve quadratically
y² + 11y + 30 = 0
y² + 6y + 5y + 30 = 0
y(y +6) +5 (y +6) = 0
(y + 5)(y + 6) = 0
y + 5 = 0 OR y + 6 = 0
y = -5 OR y = -6
Substitute the values of y into equation (3)
x = 6 + y
When y = -5
x = 6 + (-5)
x = 6 -5
x = 1
When y = -6
x = 6 + (-6)
x = 6 -6
x = 0
∴ When x = 0, y = -6 and when x = 1, y = -5
Hence, the solutions to the given system of equations is (0, -6) and (1, -5)
Learn more on Solving simultaneous equations here: brainly.com/question/16863577
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Y- mx+c
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Answer:
In order to find the slope, you need to find the difference in the x values and the difference in the y values. To do this, you write out y2-y1/x2-x1, insert your x and y values, and get your answer. Then, you find your y-intercept (where the x value is 0).
Answer:
A is 1/2
B is 2
C is -2
Step-by-step explanation:
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Answer:
The value of the side PS is 26 approx.
Step-by-step explanation:
In this question we have two right triangles. Triangle PQR and Triangle PQS.
Where S is some point on the line segment QR.
Given:
PR = 20
SR = 11
QS = 5
We know that QR = QS + SR
QR = 11 + 5
QR = 16
Now triangle PQR has one unknown side PQ which in its base.
Finding PQ:
Using Pythagoras theorem for the right angled triangle PQR.
PR² = PQ² + QR²
PQ = √(PR² - QR²)
PQ = √(20²+16²)
PQ = √656
PQ = 4√41
Now for right angled triangle PQS, PS is unknown which is actually the hypotenuse of the right angled triangle.
Finding PS:
Using Pythagoras theorem, we have:
PS² = PQ² + QS²
PS² = 656 + 25
PS² = 681
PS = 26.09
PS = 26