Question

1. given the following system of inequalities,
2. $3x - y < 5$
3. $2x - 3y > 8$

which of the following points does not satisfy the system?
(0,-2)
(-2,0)
(3,-1)
(1,3)

1. consider the system of equations shown here:

equation 1: $2x + 4y = 10$
equation 2: $5x - 8y = -9$
which of the following operations can be performed to eliminate a variable when solving using the elimination method?
(select all that apply.)
multiply equation one by 2.
multiply equation one by -2.
multiply both equations by 10.
multiply equation one by 5 and equation two by 2.

Explanation:

Sub - Question 1

Step 1: Check point (0, - 2) in \(3x - y<5\)

Substitute \(x = 0\), \(y=-2\) into \(3x - y\): \(3(0)-(-2)=0 + 2=2\). Since \(2<5\), it satisfies the first inequality.
Substitute into \(2x-3y>8\): \(2(0)-3(-2)=0 + 6 = 6\). Since \(6
ot>8\)? Wait, no, \(6<8\)? Wait, no, \(2x-3y>8\), \(2(0)-3(-2)=6\), \(6\) is not greater than \(8\)? Wait, maybe I made a mistake. Wait, the second inequality is \(2x - 3y>8\)? Wait, the original problem says \(2.2x - 3y>8\)? Wait, maybe it's a typo, maybe \(2x-3y > 8\). Let's re - check. For point (0, - 2): \(2(0)-3(-2)=6\), \(6\) is not greater than \(8\). Wait, but let's check other points.
Wait, maybe the second inequality is \(2x-3y < 8\)? No, the problem says \(2.2x - 3y>8\)? Wait, maybe it's \(2x-3y>8\). Let's check point (- 2,0):

Step 2: Check point (- 2,0) in \(3x - y<5\)

Substitute \(x=-2\), \(y = 0\): \(3(-2)-0=-6\). Since \(-6<5\), satisfies the first inequality.
Substitute into \(2x-3y>8\): \(2(-2)-3(0)=-4\). Since \(-4
ot>8\), no. Wait, maybe the second inequality is \(2x - 3y<8\)? The problem statement might have a typo. Wait, the original problem's second inequality: "2. \(2x - 3y>8\)"? Let's check point (3, - 1):

Step 3: Check point (3, - 1) in \(3x - y<5\)

Substitute \(x = 3\), \(y=-1\): \(3(3)-(-1)=9 + 1=10\). Since \(10
ot<5\), so this point does not satisfy the first inequality. Wait, let's confirm:
For \(3x - y<5\), \(3(3)-(-1)=10\), \(10\) is not less than \(5\). So we can stop here.

To eliminate a variable using the elimination method, we want to make the coefficients of one of the variables opposite (so that when we add the equations, the variable is eliminated) or equal (so that when we subtract, the variable is eliminated).
The coefficients of \(y\) are \(4\) (in Equation 1: \(2x + 4y=10\)) and \(- 8\) (in Equation 2: \(5x-8y=-9\)).

• If we multiply Equation 1 by \(2\), the coefficient of \(y\) becomes \(8\), and the coefficient of \(y\) in Equation 2 is \(-8\). Then, when we add the two equations, the \(y\) terms will be \(8y+(-8y) = 0\), and \(y\) will be eliminated.
• Multiplying Equation 1 by \(-2\) will not help in eliminating \(y\) (the coefficient of \(y\) will be \(-8\), and Equation 2 has \(-8y\), subtracting or adding will not eliminate \(y\)).
• Multiplying both equations by \(10\) does not change the coefficients' relationship, so it won't help in elimination.
• If we multiply Equation 1 by \(5\) (\(10x + 20y = 50\)) and Equation 2 by \(2\) (\(10x-16y=-18\)), the coefficients of \(x\) are both \(10\), and we can subtract the equations to eliminate \(x\).

Answer:

(3, - 1)

Sub - Question 2