Question

which system of equations has a solution of approximately (0.7, −1.4)?
○ ( x - y = 2 ) and ( x + 2y = -2 )
○ ( x - 2y = -2 ) and ( x + y = 2 )
○ ( x - y = 2 ) and ( x - 2y = -2 )
○ ( x + y = 2 ) and ( x + 2y = -2 )

Explanation:

Step1: Test the point (0.7, -1.4) in the first system: \(x - y = 2\) and \(x + 2y = -2\)

• For \(x - y = 2\): Substitute \(x = 0.7\), \(y = -1.4\). Left side: \(0.7 - (-1.4)=0.7 + 1.4 = 2.1

eq2\). So first system is out.

Step2: Test the point in the second system: \(x - 2y = -2\) and \(x + y = 2\)

• For \(x - 2y = -2\): Substitute \(x = 0.7\), \(y = -1.4\). Left side: \(0.7-2\times(-1.4)=0.7 + 2.8 = 3.5

eq - 2\). So second system is out.

Step3: Test the point in the third system: \(x - y = 2\) and \(x - 2y = -2\)

• For \(x - y = 2\): Substitute \(x = 0.7\), \(y = -1.4\). Left side: \(0.7-(-1.4)=2.1\approx2\) (close).
• For \(x - 2y = -2\): Substitute \(x = 0.7\), \(y = -1.4\). Left side: \(0.7-2\times(-1.4)=0.7 + 2.8 = 3.5

eq - 2\). Wait, maybe miscalculation. Wait, no, let's check the fourth system.

Step4: Test the point in the fourth system: \(x + y = 2\) and \(x + 2y = -2\)

• For \(x + y = 2\): Substitute \(x = 0.7\), \(y = -1.4\). Left side: \(0.7+(-1.4)=-0.7

eq2\). Wait, maybe I made a mistake in step 3. Wait, re - check the third system:
Wait, the third system is \(x - y = 2\) and \(x - 2y=-2\). Wait, no, let's re - solve the third system's equations.
From \(x - y = 2\), we have \(x=y + 2\). Substitute into \(x - 2y=-2\): \((y + 2)-2y=-2\), \(y + 2-2y=-2\), \(-y=-4\), \(y = 4\), \(x=6\). Not (0.7, - 1.4).
Wait, let's check the first system again. Wait, the first system: \(x - y = 2\) and \(x + 2y=-2\). From \(x - y = 2\), \(x=y + 2\). Substitute into \(x + 2y=-2\): \(y + 2+2y=-2\), \(3y=-4\), \(y=-\frac{4}{3}\approx - 1.33\), \(x=2-\frac{4}{3}=\frac{2}{3}\approx0.67\). Which is approximately (0.7, - 1.4). Wait, maybe the options were misread. Wait, the first option is \(x - y = 2\) and \(x + 2y=-2\). Let's calculate the values:
For \(x - y = 2\), when \(x = 0.7\), \(y=x - 2=0.7-2=-1.3\approx - 1.4\) (close). For \(x + 2y=-2\), \(0.7+2y=-2\), \(2y=-2.7\), \(y=-1.35\approx - 1.4\). So this is the closest.

Wait, maybe I made a mistake in the initial step. Let's re - evaluate each option by substituting \(x = 0.7\) and \(y=-1.4\):

Option 1: \(x - y=0.7-(-1.4)=2.1\approx2\); \(x + 2y=0.7+2\times(-1.4)=0.7-2.8=-2.1\approx - 2\).

Option 2: \(x - 2y=0.7-2\times(-1.4)=0.7 + 2.8 = 3.5
eq - 2\); \(x + y=0.7-1.4=-0.7
eq2\).

Option 3: \(x - y=0.7-(-1.4)=2.1\approx2\); \(x - 2y=0.7-2\times(-1.4)=3.5
eq - 2\).

Option 4: \(x + y=0.7-1.4=-0.7
eq2\); \(x + 2y=0.7+2\times(-1.4)=-2.1\approx - 2\).

So the first option \(x - y = 2\) and \(x + 2y=-2\) gives values closest to the given point (0.7, - 1.4).

Answer:

A. \(x - y = 2\) and \(x + 2y=-2\)