Which point would be a solution to the system of linear inequalities shown below?
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)
11 answers
To determine which point is a solution to the system of inequalities, we need to test each point in the inequalities given:
The only point that satisfies both inequalities is \((-10, -6)\).
Thus, the solution to the system of inequalities is
\[
\boxed{(-10, -6)}
\]
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)
To determine which point satisfies both inequalities, we need to analyze each inequality and see where the solution regions are located:
Inequality 1: \( y > -\frac{1}{2}x - 2 \)
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.
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:
\[
\boxed{(-2, 1)}
\]
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.
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:
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.
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 \( x \).
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.
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:
\[
\text{A, add to eliminate } y. \text{ Add to eliminate } y.
\]
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
