Which system of equations has a solution of approximately $$ (0.7,-1.4) $$ ? $$ \begin{array}{l}x-y=2 ext { and } x+2 y=-2 \ x-2 y=-2 ext { and } x+y=2 \ x-y=2 ext { and } x-2 y=-2 \ x+y=2 ext { and } x+2 y=-2\end{array} $$ Mark this and return

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Which system of equations has a solution of approximately $$ (0.7,-1.4) $$ ? $$ \begin{array}{l}x-y=2 	ext { and } x+2 y=-2 \ x-2 y=-2 	ext { and } x+y=2 \ x-y=2 	ext { and } x-2 y=-2 \ x+y=2 	ext { and } x+2 y=-2\end{array} $$ Mark this and return

UpStudy AI · Accepted Answer

We start by examining each system of equations to see if it has an approximate solution of $$ (0.7, -1.4) $$.

### Option 1
The system is
$$
x - y = 2 \quad \text{and} \quad x + 2y = -2.
$$
1. Solve the first equation for $$ y $$:
   $$
   y = x - 2.
   $$
2. Substitute into the second equation:
   $$
   x + 2(x - 2) = -2 \quad \Longrightarrow \quad x + 2x - 4 = -2.
   $$
3. Combine like terms:
   $$
   3x - 4 = -2.
   $$
4. Solve for $$ x $$:
   $$
   3x = 2 \quad \Longrightarrow \quad x = \frac{2}{3} \approx 0.67.
   $$
5. Substitute $$ x $$ back to find $$ y $$:
   $$
   y = \frac{2}{3} - 2 = -\frac{4}{3} \approx -1.33.
   $$

The approximate solution is $$ (0.67, -1.33) $$, which is very close to $$ (0.7, -1.4) $$.

### Option 2
The system is
$$
x - 2y = -2 \quad \text{and} \quad x + y = 2.
$$
1. Solve the second equation for $$ x $$:
   $$
   x = 2 - y.
   $$
2. Substitute into the first equation:
   $$
   (2 - y) - 2y = -2 \quad \Longrightarrow \quad 2 - 3y = -2.
   $$
3. Solve for $$ y $$:
   $$
   -3y = -4 \quad \Longrightarrow \quad y \approx 1.33.
   $$
4. Thus, $$ x = 2 - 1.33 \approx 0.67 $$.

This gives the solution $$ (0.67, 1.33) $$, which does not match $$ (0.7, -1.4) $$.

### Option 3
The system is
$$
x - y = 2 \quad \text{and} \quad x - 2y = -2.
$$
1. From the first equation, $$ x = y + 2 $$.
2. Substitute into the second:
   $$
   (y + 2) - 2y = -2 \quad \Longrightarrow \quad 2 - y = -2.
   $$
3. Solve for $$ y $$:
   $$
   -y = -4 \quad \Longrightarrow \quad y = 4.
   $$
4. Then, $$ x = 4 + 2 = 6 $$.

The solution here is $$ (6, 4) $$, which is not an approximate match.

### Option 4
The system is
$$
x + y = 2 \quad \text{and} \quad x + 2y = -2.
$$
1. Subtract the first equation from the second:
   $$
   (x + 2y) - (x + y) = -2 - 2 \quad \Longrightarrow \quad y = -4.
   $$
2. Substitute $$ y = -4 $$ into $$ x + y = 2 $$:
   $$
   x - 4 = 2 \quad \Longrightarrow \quad x = 6.
   $$

This gives the solution $$ (6, -4) $$, which is not an approximate match.

### Conclusion
Only **Option 1** produces a solution that is approximately $$ (0.7, -1.4) $$.

$$
\boxed{x - y = 2 \quad \text{and} \quad x + 2y = -2.}
$$
The system $$ x - y = 2 $$ and $$ x + 2y = -2 $$ has an approximate solution of $$ (0.7, -1.4) $$.

Which system of equations has a solution of approximately \( (0.7,-1.4) \) ? \( \begin{array}{l}x-y=2 \text { and } x+2 y=-2 \\ x-2 y=-2 \text { and } x+y=2 \\ x-y=2 \text { and } x-2 y=-2 \\ x+y=2 \text { and } x+2 y=-2\end{array} \) Mark this and return

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