Solve the system of equations
y=3x
x^2-y=28
Substitute 3x for y in the second equation and solve for x.
x^2 - 3x = 28
x(x-3) = 28
x = either 28 or 25
Use these value to solve for y in the first equation. Check by inserting values into the second equation.
To solve the system of equations, we need to find the values of x and y that satisfy both equations simultaneously. Let's solve this step by step:
First, let's rewrite the first equation in terms of y:
y = 3x
Next, substitute this expression for y into the second equation:
x^2 - (3x) = 28
Rearrange this equation to bring all the terms to one side:
x^2 - 3x - 28 = 0
We now have a quadratic equation, which we can solve by factoring, completing the square, or using the quadratic formula.
To factor this quadratic equation, we need to find two numbers that multiply to -28 and add up to -3. After some trial and error, we can factor the quadratic equation as:
(x - 7)(x + 4) = 0
Now, we can set each factor equal to zero and solve for x:
x - 7 = 0 or x + 4 = 0
Solving these equations, we find:
x = 7 or x = -4
Since we have two possible values for x, we can substitute each value back into either equation to find the corresponding values for y.
For x = 7:
y = 3x
y = 3(7)
y = 21
So one solution to the system of equations is x = 7, y = 21.
For x = -4:
y = 3x
y = 3(-4)
y = -12
So another solution to the system of equations is x = -4, y = -12.
Therefore, the system of equations has two solutions: (7, 21) and (-4, -12).