Let's solve the system of equations:

1. $y = x^2$
2. $7x + 2y = 4$

Step 1: Substitute $y$ from the first equation into the second equation.

From equation (1), we know that $y = x^2$. Now substitute this into equation (2):

$7x + 2(x^2) = 4$

Step 2: Simplify and solve the resulting quadratic equation.

$7x + 2x^2 = 4$

Rearrange the terms to form a standard quadratic equation:

$2x^2 + 7x - 4 = 0$

Step 3: Solve the quadratic equation using the quadratic formula.

The quadratic formula is given by:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

For the equation $2x^2 + 7x - 4 = 0$, the coefficients are $a = 2$, $b = 7$, and $c = -4$. Substituting these into the quadratic formula:

$x = \frac{-7 \pm \sqrt{7^2 - 4(2)(-4)}}{2(2)}$ $x = \frac{-7 \pm \sqrt{49 + 32}}{4}$ $x = \frac{-7 \pm \sqrt{81}}{4}$ $x = \frac{-7 \pm 9}{4}$

Thus, we have two possible solutions for $x$:

1. $x = \frac{-7 + 9}{4} = \frac{2}{4} = 0.5$
2. $x = \frac{-7 - 9}{4} = \frac{-16}{4} = -4$

Step 4: Find corresponding $y$-values.

Using the first equation $y = x^2$, we can now find the corresponding $y$-values for each $x$-solution:

• For $x = 0.5$, $y = (0.5)^2 = 0.25$
• For $x = -4$, $y = (-4)^2 = 16$

Final Solutions:

The solutions to the system of equations are:

• $x = 0.5, y = 0.25$
• $x = -4, y = 16$

Do you need further details on the solution or have additional questions?

Related Questions:

1. How do you use the quadratic formula to solve quadratic equations?
2. What are the real-world applications of quadratic equations?
3. Can you solve a system of equations involving linear and quadratic functions graphically?
4. How do you check the solutions of a system of equations?
5. How do you approach a system with more than two equations?

Tip:

When solving a system of equations involving a quadratic and linear equation, substitution is a great method to simplify the problem.