To solve the system of equations by substitution, we can set the two equations equal to each other:
x^2 - 3x - 7 = -2x^2 - 4x - 3
Now, let's simplify this equation:
3x^2 + x - 4 = 0
To solve this quadratic equation, we can factor it as follows:
(3x - 4)(x + 1) = 0
Setting each factor equal to zero, we get:
3x - 4 = 0 or x + 1 = 0
Solving for x in each equation, we have:
3x = 4 or x = -1
Dividing both sides by 3 in the first equation gives us:
x = 4/3
So the two possible values for x are x = 4/3 or x = -1.
Now, we can substitute these values back into one of the original equations to solve for y. Let's use the second equation:
y = -2x^2 - 4x - 3
For x = 4/3:
y = -2(4/3)^2 - 4(4/3) - 3
= -2(16/9) - 16/3 - 3
= -32/9 - 48/9 - 27/9
= (-32 - 48 - 27)/9
= -107/9
So for x = 4/3, y = -107/9.
And for x = -1:
y = -2(-1)^2 - 4(-1) - 3
= -2(1) + 4 - 3
= -2 + 4 - 3
= -1
So for x = -1, y = -1.
Therefore, the solutions to the system of equations are (4/3, -107/9) and (-1, -1).
Solve the system of equations by substitution. (Simplify your answer. Use integers or fractions for any numbers in the expression. Type an ordered pair. Use a comma to separate answers as needed)
y = x^2 - 3x - 7
y = -2x^2 - 4x - 3
1 answer