Question

14x−4y=40
7x + 8 y= -10
solve by elimanation

Answer 1

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

1. \( 14x - 4y = 40 \) (Equation 1)
2. \( 7x + 8y = -10 \) (Equation 2)

**Step 1: Make the coefficients of \(y\) in both equations equal.**

We can multiply Equation 1 by 2 to get the coefficients of \(y\) to match:

\[
2(14x - 4y) = 2(40)
\]

This gives us:

\[
28x - 8y = 80 \quad \text{(Equation 3)}
\]

Now we have:

1. \( 28x - 8y = 80 \) (Equation 3)
2. \( 7x + 8y = -10 \) (Equation 2)

**Step 2: Add Equation 3 and Equation 2.**

We will add the two equations together to eliminate \(y\):

\[
(28x - 8y) + (7x + 8y) = 80 - 10
\]

This simplifies to:

\[
35x + 0y = 70
\]

So we have:

\[
35x = 70
\]

**Step 3: Solve for \(x\).**

Divide both sides by 35:

\[
x = \frac{70}{35} = 2
\]

**Step 4: Substitute \(x = 2\) back into one of the original equations.**

Let's use Equation 2:

\[
7x + 8y = -10
\]

Substituting for \(x\):

\[
7(2) + 8y = -10
\]

This simplifies to:

\[
14 + 8y = -10
\]

Subtract 14 from both sides:

\[
8y = -10 - 14
\]
\[
8y = -24
\]

Now, divide by 8:

\[
y = \frac{-24}{8} = -3
\]

**Final Solution:**

The solution for the system of equations is:

\[
x = 2, \quad y = -3
\]

Thus, the point of intersection is \( (2, -3) \).