How to solve quadratic equations with the coefficient not equal to 1?
1 answer:
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Answer:
The ordered pair is indeed a solution to the system:
.
Step-by-step explanation:
Consider a system of equations about variables and
. An ordered pair
(where
and
are constant) is a solution to that system if and only if all equations in that system hold after substituting in
and
.
For the system in this question, would be a solution only if both equations in the system hold after replacing all
in equations of the system with
and all
with
.
The of the equation
would become
. The
of that equation would become
. The two sides are indeed equal.
Similarly, the of the equation
would become
. The
of that equation would become
. The two sides are indeed equal.
Thus, and
simultaneously satisfy both equations of the given system. Therefore, the ordered pair
would indeed be a solution to that system.
Answer:
(-2.2, -1.6), (3, 1)
Step-by-step explanation:
You don't have to go far to find the equations. They are right there in your problem statement. Perhaps you want to find the solutions to the equations.
Use the first equation to write an expression for x, then substitute that into the second equation:
x = 2y +1
y^2 -3(2y+1)(y) +8 = 0
-5y^2 -3y +8 = 0
-(5y +8)(y -1) = 0
y = -8/5 or y = 1
The corresponding values of x are ...
x = 2(-8/5)+1 = -11/5
x = 2(1) +1 = 3
The solutions are (x, y) = (-2.2, -1.6) and (3, 1).
