Question

Consider the following system of equations:

{
3
1
​
x+y=3
2y=−
3
2
​
x+10
​

0
/
2
11.
Is
(0,3) a solution to the system of equations? Explain.


no because it not a solution in the system of the eqations
0
/
2
12.
Solve the system of equations using the elimination method. Make sure to explain what you are doing.

Answer 1

Let's break down the system of equations you've provided:

1. \( x + y = 3 \) (Equation 1)
2. \( 2y = -\frac{3}{2}x + 10 \) (Equation 2)

### Step 1: Check if (0, 3) is a solution

We need to substitute \( x = 0 \) and \( y = 3 \) into both equations to see if they hold true.

#### Checking Equation 1:
\[
0 + 3 = 3
\]
This holds true.

#### Checking Equation 2:
Substituting \( x = 0 \) and \( y = 3 \):
\[
2(3) = -\frac{3}{2}(0) + 10
\]
This simplifies to:
\[
6 = 0 + 10
\]
Which is false. Therefore, (0, 3) is **not** a solution to the system of equations.

---

### Step 2: Solve the system using the elimination method

To solve the system, we will manipulate the equations so that we can eliminate one variable.

**Step 2.1: Rewrite Equation 2**

First, let's simplify Equation 2:
\[
2y = -\frac{3}{2}x + 10
\]
We can divide the entire equation by 2 to obtain:
\[
y = -\frac{3}{4}x + 5
\]
(We can also use the original form in elimination, but for clarity, we'll continue with the transformed Equation 2.)

**Step 2.2: Eliminate one variable**

Now we will use Equation 1, \( x + y = 3 \), and substitute for \( y \) from the modified Equation 2:

Substituting \( y \):
\[
x + \left(-\frac{3}{4}x + 5\right) = 3
\]
Combine like terms:
\[
x - \frac{3}{4}x + 5 = 3
\]

Now, combine \( x \) and \(-\frac{3}{4}x\):
\[
\frac{1}{4}x + 5 = 3
\]

**Step 2.3: Isolate \( x \)**

Subtract 5 from both sides:
\[
\frac{1}{4}x = 3 - 5
\]
\[
\frac{1}{4}x = -2
\]

Multiply by 4 to solve for \( x \):
\[
x = -2 \cdot 4
\]
\[
x = -8
\]

**Step 2.4: Substitute back to find \( y \)**

Now substitute \( x = -8 \) back into Equation 1 to solve for \( y \):
\[
-8 + y = 3
\]
Add 8 to both sides:
\[
y = 3 + 8
\]
\[
y = 11
\]

### Solution

The solution to the system of equations is:
\[
(x, y) = (-8, 11)
\]

### Final Verification
Let's quickly verify by substituting \( (-8, 11) \) into both original equations:

1. **For Equation 1**:
\[
-8 + 11 = 3 \quad \text{(True)}
\]

2. **For Equation 2**:
Substituting in the original form:
\[
2(11) = -\frac{3}{2}(-8) + 10
\]
Calculating left side:
\[
22
\]
Calculating right side:
\[
12 + 10 = 22 \quad \text{(True)}
\]

Both equations are satisfied, confirming that our solution \( (-8, 11) \) is correct.