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solve the following system of equations algebraically

y=x^2-5x+4
y=x+4

1 answer

To solve the system of equations algebraically, we have the following two equations:

  1. \( y = x^2 - 5x + 4 \) (Equation 1)
  2. \( y = x + 4 \) (Equation 2)

Since both equations equal \( y \), we can set them equal to each other:

\[ x^2 - 5x + 4 = x + 4 \]

Now, we can rearrange the equation by moving all terms to one side:

\[ x^2 - 5x + 4 - x - 4 = 0 \]

This simplifies to:

\[ x^2 - 6x = 0 \]

We can factor out an \( x \):

\[ x(x - 6) = 0 \]

Setting each factor to zero gives us the possible solutions for \( x \):

  1. \( x = 0 \)
  2. \( x - 6 = 0 \) implies \( x = 6 \)

Now, we will find the corresponding \( y \) values by substituting \( x \) back into Equation 2:

  1. For \( x = 0 \): \[ y = 0 + 4 = 4 \] So, one solution is \( (0, 4) \).

  2. For \( x = 6 \): \[ y = 6 + 4 = 10 \] So, the other solution is \( (6, 10) \).

Thus, the solutions to the system of equations are:

\[ (0, 4) \quad \text{and} \quad (6, 10) \]