15. What is the solution of the linear-quadratic system of equations?
y=x^2+5x-3
y-x=2
how about
x^2+5x-3 = x+2
x^2+4x-5 = 0
Now it's a cinch, right?
To find the solution of a linear-quadratic system of equations, we need to solve the system by substitution or elimination. Let's solve it by substitution.
We are given the following system of equations:
y = x^2 + 5x - 3 -- (Equation 1)
y - x = 2 -- (Equation 2)
Step 1: Solve Equation 2 for y.
Rearrange Equation 2 to get: y = x + 2
Step 2: Substitute this value of y in Equation 1.
Replace y in Equation 1 with x + 2, so we have:
x + 2 = x^2 + 5x - 3
Step 3: Rearrange the equation to make it a quadratic equation equal to zero.
Move all terms to one side:
x^2 + 5x - 3 - x - 2 = 0
x^2 + 4x - 5 = 0
Step 4: Solve the quadratic equation.
We can solve this equation by factoring, completing the square, or using the quadratic formula.
Factoring is the most straightforward method in this case:
(x-1)(x+5) = 0
From this factorization, we get two solutions:
x - 1 = 0, which gives x = 1
x + 5 = 0, which gives x = -5
Step 5: Substitute the values of x back into Equation 2 to find the corresponding y-values.
Substituting x = 1 into Equation 2:
y = 1 + 2 = 3
So one solution is (x, y) = (1, 3).
Substituting x = -5 into Equation 2:
y = -5 + 2 = -3
The other solution is (x, y) = (-5, -3).
Therefore, the solution to the given system of equations is (1, 3) and (-5, -3).