Question

Solve the following system of equations using the elimination method. 2x−3y=14 , 4x+3y=10 (1 point) ((), ())

Answer 1

Use the elimination method to solve the linear-quadratic system y=x^2+2x−4 , y=−x−6 .(1 point) The solutions are (−1,−5) and ().

Answer 2

To solve the system of equations using the elimination method, we have the following equations:

1) \( 2x - 3y = 14 \)
2) \( 4x + 3y = 10 \)

We can eliminate \( y \) by adding the two equations together. First, let's manipulate the first equation to make the coefficients of \( y \) in both equations match in magnitude but opposite in sign.

Multiply the first equation by 1, and leave the second as is:

1) \( 2x - 3y = 14 \)
2) \( 4x + 3y = 10 \)

Now, let's add both equations:

\[
(2x - 3y) + (4x + 3y) = 14 + 10
\]

This simplifies to:

\[
(2x + 4x) + (-3y + 3y) = 24
\]

\[
6x = 24
\]

Now, solve for \( x \):

\[
x = \frac{24}{6} = 4
\]

Next, substitute \( x = 4 \) back into one of the original equations to find \( y \). We'll use the first equation:

\[
2(4) - 3y = 14
\]

This simplifies to:

\[
8 - 3y = 14
\]

Now isolate \( y \):

\[
-3y = 14 - 8
\]
\[
-3y = 6
\]
\[
y = \frac{6}{-3} = -2
\]

Thus, the solution to the system of equations is:

\[
(x, y) = (4, -2)
\]

So, the final answer is:

\[
(4, -2)
\]

Answer 3

To solve the linear-quadratic system given by:

1) \( y = x^2 + 2x - 4 \)
2) \( y = -x - 6 \)

we can substitute the expression for \( y \) from the linear equation into the quadratic equation.

Set the two equations equal to each other:

\[
-x - 6 = x^2 + 2x - 4
\]

Now, rearrange the equation to form a standard quadratic equation:

\[
0 = x^2 + 2x - 4 + x + 6
\]

Combine like terms:

\[
0 = x^2 + 3x + 2
\]

Now, we can factor the quadratic:

\[
0 = (x + 1)(x + 2)
\]

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

1) \( x + 1 = 0 \) → \( x = -1 \)
2) \( x + 2 = 0 \) → \( x = -2 \)

Now that we have the \( x \)-values, we can substitute these back into one of the original equations to find the corresponding \( y \)-values. We'll use the linear equation \( y = -x - 6 \).

1) For \( x = -1 \):

\[
y = -(-1) - 6 = 1 - 6 = -5
\]

So, one solution is \( (-1, -5) \).

2) For \( x = -2 \):

\[
y = -(-2) - 6 = 2 - 6 = -4
\]

Thus, the second solution is \( (-2, -4) \).

Therefore, the solutions to the system are:

\[
(-1, -5) \quad \text{and} \quad (-2, -4)
\]

To summarize, the solutions are \((-1, -5)\) and \((-2, -4)\).