Which point would be a solution to the system of linear inequalities shown below?

Question

y, is less than or equal to, minus, one fifth, x, minus, 8, y, is greater than or equal to, 5, x, plus, 5
y≤−
5
1
​
x−8y≥5x+5
Answer
Multiple Choice Answers
left parenthesis, minus, 5, comma, 3, right parenthesis(−5,3)
left parenthesis, minus, 10, comma, 5, right parenthesis(−10,5)
left parenthesis, minus, 10, comma, minus, 6, right parenthesis(−10,−6)
left parenthesis, minus, 5, comma, 2, right parenthesis(−5,2)

Answer 1

To determine which point is a solution to the system of inequalities, we need to test each point in the inequalities given:

1. Inequality 1: \( y \leq -\frac{1}{5}x - 8 \)
2. Inequality 2: \( y \geq 5x + 5 \)

Let's evaluate each point:

Point 1: \((-5, 3)\)

1. For \( y \leq -\frac{1}{5}x - 8 \): \[ 3 \leq -\frac{1}{5}(-5) - 8 \ 3 \leq 1 - 8 \ 3 \leq -7 \quad \text{(false)} \]

This point does not satisfy the first inequality.

Point 2: \((-10, 5)\)

1. For \( y \leq -\frac{1}{5}x - 8 \): \[ 5 \leq -\frac{1}{5}(-10) - 8 \ 5 \leq 2 - 8 \ 5 \leq -6 \quad \text{(false)} \]

This point does not satisfy the first inequality.

Point 3: \((-10, -6)\)

1. For \( y \leq -\frac{1}{5}x - 8 \): \[ -6 \leq -\frac{1}{5}(-10) - 8 \ -6 \leq 2 - 8 \ -6 \leq -6 \quad \text{(true)} \]

2. For \( y \geq 5x + 5 \): \[ -6 \geq 5(-10) + 5 \ -6 \geq -50 + 5 \ -6 \geq -45 \quad \text{(true)} \]

This point satisfies both inequalities.

Point 4: \((-5, 2)\)

1. For \( y \leq -\frac{1}{5}x - 8 \): \[ 2 \leq -\frac{1}{5}(-5) - 8 \ 2 \leq 1 - 8 \ 2 \leq -7 \quad \text{(false)} \]

This point does not satisfy the first inequality.

Conclusion:

The only point that satisfies both inequalities is \((-10, -6)\).

Thus, the solution to the system of inequalities is

\[ \boxed{(-10, -6)} \]

Answer 2

Below are two inequalities and the graphs of their lines without the shading. By imagining where the shading should be, identify which point would satisfy BOTH inequalities.
y, is greater than, minus, one half, x, minus, 2
y>−
2
1
​
x−2
y, is greater than, x, plus, 1
y>x+1
x
y
Answer
Multiple Choice Answers
left parenthesis, minus, 6, comma, minus, 4, right parenthesis(−6,−4)
left parenthesis, minus, 2, comma, 1, right parenthesis(−2,1)
left parenthesis, 8, comma, 8, right parenthesis(8,8)
left parenthesis, 0, comma, minus, 7, right parenthesis(0,−7)

Answer 3

To determine which point satisfies both inequalities, we need to analyze each inequality and see where the solution regions are located:

1. Inequality 1: \( y > -\frac{1}{2}x - 2 \)
2. Inequality 2: \( y > x + 1 \)

Step 1: Determine the regions for each inequality

• For the first inequality, the line \( y = -\frac{1}{2}x - 2 \) has a downward slope and y-intercept of -2. The region satisfying this inequality is above this line.

• For the second inequality, the line \( y = x + 1 \) has an upward slope of 1 and y-intercept of 1. The region satisfying this inequality is also above this line.

Step 2: Verify each point

Now, we will substitute each point into both inequalities to see if they satisfy both conditions.

Point 1: \((-6, -4)\)

1. For \( y > -\frac{1}{2}x - 2 \): \[ -4 > -\frac{1}{2}(-6) - 2 \ -4 > 3 - 2 \ -4 > 1 \quad \text{(false)} \]

2. For \( y > x + 1 \): \[ -4 > -6 + 1 \ -4 > -5 \quad \text{(true)} \]

Point \((-6, -4)\) does not satisfy the first inequality.

Point 2: \((-2, 1)\)

1. For \( y > -\frac{1}{2}x - 2 \): \[ 1 > -\frac{1}{2}(-2) - 2 \ 1 > 1 - 2 \ 1 > -1 \quad \text{(true)} \]

2. For \( y > x + 1 \): \[ 1 > -2 + 1 \ 1 > -1 \quad \text{(true)} \]

Point \((-2, 1)\) satisfies both inequalities.

Point 3: \((8, 8)\)

1. For \( y > -\frac{1}{2}x - 2 \): \[ 8 > -\frac{1}{2}(8) - 2 \ 8 > -4 - 2 \ 8 > -6 \quad \text{(true)} \]

2. For \( y > x + 1 \): \[ 8 > 8 + 1 \ 8 > 9 \quad \text{(false)} \]

Point \((8, 8)\) does not satisfy the second inequality.

Point 4: \((0, -7)\)

1. For \( y > -\frac{1}{2}x - 2 \): \[ -7 > -\frac{1}{2}(0) - 2 \ -7 > -2 \quad \text{(false)} \]

2. For \( y > x + 1 \): \[ -7 > 0 + 1 \ -7 > 1 \quad \text{(false)} \]

Point \((0, -7)\) does not satisfy either inequality.

Conclusion

The only point that satisfies both inequalities is \((-2, 1)\).

Thus, the solution is:

\[ \boxed{(-2, 1)} \]

Answer 4

A variable needs to be eliminated to solve the system of equations below. Choose the correct first step.
5, x, plus, y, equals, 48
5x+y=
48
3, x, minus, y, equals, 16
3x−y=
16
Answer
Multiple Choice Answers
A, d, d, to eliminate , y, .Add to eliminate y.
S, u, b, t, r, a, c, t, to eliminate , x, .Subtract to eliminate x.
S, u, b, t, r, a, c, t, to eliminate , y, .Subtract to eliminate y.
A, d, d, to eliminate , x, .Add to eliminate x.

Answer 5

To solve the system of equations

\[ 5x + y = 48 \] \[ 3x - y = 16 \]

we can choose to eliminate either variable \( x \) or \( y \). Let's analyze the options:

1. Option A: "Add to eliminate \( y \)."

   • If we add the two equations, the \( y \) terms would cancel out because \( y \) appears as \( +y \) in the first equation and \( -y \) in the second equation.

2. Option B: "Subtract to eliminate \( x \)."

   • If we were to subtract the first equation from the second (or vice versa), the \( x \) terms would not cancel out nicely because they both involve \( x \) with different coefficients. This is not the correct approach for eliminating \( x \).

3. Option C: "Subtract to eliminate \( y \)."

   • If we were to subtract the second equation from the first, we wouldn't eliminate \( y \) because the signs in front of \( y \) in the equations are \( +y \) and \( -y \). This would not give us a cancellation.

4. Option D: "Add to eliminate \( x \)."

   • Adding the equations would not eliminate \( x \) as both equations have different coefficients for \( x \).

Conclusion

The correct first step is:

• Add to eliminate \( y \) (Option A).

Thus, the answer is:

\[ \text{A, add to eliminate } y. \text{ Add to eliminate } y. \]

Answer 6

nd the solution of the system of equations.
minus, 10, x, minus, y, equals, minus, 29
−10x−y=
−29
10, x, plus, 7, y, equals, 23
10x+7y=
23
Answer
Attempt 1 out of 2

left parenthesis(
,
right parenthesis) Submit Answer

Answer 7

To solve the system of equations

\[ -10x - y = -29 \quad \text{(1)} \] \[ 10x + 7y = 23 \quad \text{(2)} \]

we can use the method of elimination or substitution. Here, we'll use the elimination method.

Step 1: Eliminate \(y\)

From equation (1), we can express \(y\) in terms of \(x\):

\[ -y = -29 + 10x \ y = 10x + 29 \quad \text{(3)} \]

Step 2: Substitute into the second equation

Now, we will substitute equation (3) into equation (2):

\[ 10x + 7(10x + 29) = 23 \ 10x + 70x + 203 = 23 \ 80x + 203 = 23 \]

Step 3: Solve for \(x\)

Now, isolate \(x\):

\[ 80x = 23 - 203 \ 80x = -180 \ x = -\frac{180}{80} \ x = -\frac{9}{4} \quad \text{or } -2.25 \]

Step 4: Substitute \(x\) back to find \(y\)

Now, substitute \(x = -\frac{9}{4}\) back into equation (3):

\[ y = 10\left(-\frac{9}{4}\right) + 29 \ y = -\frac{90}{4} + 29 \ y = -\frac{90}{4} + \frac{116}{4} \ y = \frac{26}{4} \ y = \frac{13}{2} \quad \text{or } 6.5 \]

Conclusion

The solution to the system of equations is:

\[ \left(-\frac{9}{4}, \frac{13}{2}\right) \quad \text{or } \left(-2.25, 6.5\right) \]

So, the final answer is:

\[ \boxed{\left(-\frac{9}{4}, \frac{13}{2}\right)} \] or \[ \boxed{(-2.25, 6.5)} \]

Please choose the format that’s preferred based on your submission requirements.

Answer 8

Find the solution of the system of equations.
minus, 2, x, minus, 2, y, equals, 0
−2x−2y=
0
minus, x, plus, 4, y, equals, 40
−x+4y=
40

Answer 9

To solve the system of equations

\[ -2x - 2y = 0 \quad \text{(1)} \] \[ -x + 4y = 40 \quad \text{(2)} \]

we can simplify and solve them step by step.

Step 1: Simplify Equation (1)

From equation (1), we can simplify it:

\[ -2x - 2y = 0 \ x + y = 0 \quad \text{(3)} \]

This means

\[ y = -x \]

Step 2: Substitute Equation (3) into Equation (2)

Now, we'll substitute equation (3) into equation (2):

\[ -x + 4(-x) = 40 \ -x - 4x = 40 \ -5x = 40 \]

Step 3: Solve for \(x\)

Now we will solve for \(x\):

\[ x = -\frac{40}{5} \ x = -8 \]

Step 4: Substitute back to find \(y\)

Now we can substitute \(x = -8\) back into equation (3) to find \(y\):

\[ y = -(-8) = 8 \]

Conclusion

The solution to the system of equations is:

\[ \boxed{(-8, 8)} \]

So the final answer is \((-8, 8)\).

Answer 10

Solve the system by substitution.
y, equals, 3, x, plus, 14
y=
3x+14
y, equals, x
y=
x

Answer 11

To solve the system of equations using substitution, we have:

1. \( y = 3x + 14 \) \quad (1)
2. \( y = x \) \quad (2)

Step 1: Substitute Equation (2) into Equation (1)

Since both equations are equal to \( y \), we can set them equal to each other:

\[ 3x + 14 = x \]

Step 2: Solve for \( x \)

Now, we will isolate \( x \):

\[ 3x + 14 = x \ 3x - x + 14 = 0 \ 2x + 14 = 0 \ 2x = -14 \ x = -7 \]

Step 3: Substitute \( x \) back into either equation to find \( y \)

Now that we have found \( x \), we can substitute \( x = -7 \) back into equation (2):

\[ y = x \ y = -7 \]

Conclusion

The solution to the system is:

\[ \boxed{(-7, -7)} \]

So the final answer is \( (-7, -7) \).