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Why is the system of equations 7x+7y=42

and 6x+y=36
equivalent to the system of equations x+y=6
and 6x+y=36
?(1 point)
Responses

The equation 7x+7y=42
can be multiplied by 7 to produce the equivalent equation x+y=6
.
The equation 7 x plus 7 y is equal to 42 can be multiplied by 7 to produce the equivalent equation x plus y is equal to 6. - incorrect

The intercepts of the equations 7x+7y=42
and x+y=6
are (0,36)
and (6,0)
.
The intercepts of the equations 7 x plus 7 y is equal to 42 and x plus y is equal to 6 are open paren 0 comma 36 close paren and open paren 6 comma 0 close paren. - no response given

All equations share the point (6,0)
.
All equations share the point open paren 6 comma 0 close paren. - no response given

The intercepts of the equations 7x+7y=42
and x+y=6
are (0,6)
and (6,0)
.

7 answers

To understand why the system of equations \( 7x + 7y = 42 \) and \( 6x + y = 36 \) is equivalent to the system \( x + y = 6 \) and \( 6x + y = 36 \), let's focus on the first equation.

Simplifying the First Equation

  1. Take the equation \( 7x + 7y = 42 \).
  2. You can divide every term in the equation by 7:

\[ \frac{7x}{7} + \frac{7y}{7} = \frac{42}{7} \]

This simplifies to:

\[ x + y = 6 \]

Thus, we see that \( 7x + 7y = 42 \) is equivalent to \( x + y = 6 \).

Understanding the Second Equation

The second equation in both systems is \( 6x + y = 36 \), which remains unchanged in both systems.

Conclusion on Equivalence

Therefore, the system of equations:

  1. \( 7x + 7y = 42 \)
  2. \( 6x + y = 36 \)

can be reduced to:

  1. \( x + y = 6 \)
  2. \( 6x + y = 36 \)

Both systems essentially describe the same relationship between \( x \) and \( y \).

Correct Response

The correct response from your options should articulate that dividing the first equation by 7 yields \( x + y = 6 \), which maintains the equivalence of the two systems. A full statement could be:

"The equation \( 7x + 7y = 42 \) can be divided by 7 to produce the equivalent equation \( x + y = 6 \)."

So, the second response option stating this indicates an incorrect multiplication. Is there a specific response you'd like to select or elaborate on?

Solve the following system of equations using the elimination method.

0.1x+0.3y=0.9
, 0.1x−0.3y=0.2


(1 point)
Responses

There is no solution
There is no solution

(11/2,7/6)
left parenthesis Start Fraction 11 over 2 End Fraction comma Start Fraction 7 over 6 End Fraction right parenthesis

(11/2,−7/6)
left parenthesis Start Fraction 11 over 2 End Fraction comma negative Start Fraction 7 over 6 End Fraction right parenthesis

(−7/6,11/2)
left parenthesis negative Start Fraction 7 over 6 End Fraction comma Start Fraction 11 over 2 End Fraction right parenthesis

To solve the system of equations using the elimination method, we start with the given equations:

  1. \( 0.1x + 0.3y = 0.9 \)
  2. \( 0.1x - 0.3y = 0.2 \)

Step 1: Align the equations

We can write them down clearly for elimination:

  1. \( 0.1x + 0.3y = 0.9 \) (equation 1)
  2. \( 0.1x - 0.3y = 0.2 \) (equation 2)

Step 2: Eliminate \(x\)

To eliminate \(x\), we can subtract equation 2 from equation 1:

\[ (0.1x + 0.3y) - (0.1x - 0.3y) = 0.9 - 0.2 \]

Step 3: Simplify the left side

This can be simplified to:

\[ 0.1x + 0.3y - 0.1x + 0.3y = 0.7 \]

Which simplifies to:

\[ 0.6y = 0.7 \]

Step 4: Solve for \(y\)

Now, divide both sides by \(0.6\):

\[ y = \frac{0.7}{0.6} = \frac{7}{6} \]

Step 5: Substitute \(y\) back into one of the equations to find \(x\)

We can substitute \(y = \frac{7}{6}\) into equation 1:

\[ 0.1x + 0.3\left(\frac{7}{6}\right) = 0.9 \]

Calculating \(0.3 \times \frac{7}{6}\):

\[ 0.1x + \frac{21}{60} = 0.9 \quad \text{(since } 0.3 = \frac{3}{10} = \frac{18}{60}) \]

Subtract \(\frac{21}{60}\) from both sides:

\[ 0.1x = 0.9 - \frac{21}{60} = \frac{54}{60} - \frac{21}{60} = \frac{33}{60} \]

Step 6: Solve for \(x\)

Now, divide both sides by \(0.1\):

\[ x = \frac{33/60}{0.1} = \frac{33/60}{1/10} = \frac{33}{60} \times 10 = \frac{33}{6} = \frac{11}{2} \]

Final Solution

Thus, we have:

\[ (x, y) = \left( \frac{11}{2}, \frac{7}{6} \right) \]

The correct answer is:

\(\left(\frac{11}{2}, \frac{7}{6}\right)\).

Use the image to answer the question.

Two dashed lines and four points are graphed on a coordinate plane. The x-axis ranges from negative 5 to 15 in increments of 1. The y-axis ranges from negative 10 to 12 in increments of 1.

Which point is a solution for the simultaneous inequalities y>−2x+3
and y<3x−3
?

(1 point)
Responses

(2,5)
left parenthesis 2 comma 5 right parenthesis

(3,6)
left parenthesis 3 comma 6 right parenthesis

(10,5)
left parenthesis 10 comma 5 right parenthesis

(1,−5)

To determine which point is a solution for the simultaneous inequalities \( y > -2x + 3 \) and \( y < 3x - 3 \), we can evaluate each potential point by substituting the coordinates into both inequalities.

Step 1: Evaluate the inequalities

Inequality 1: \( y > -2x + 3 \)

Inequality 2: \( y < 3x - 3 \)

Given Points:

  1. \( (2, 5) \)
  2. \( (3, 6) \)
  3. \( (10, 5) \)
  4. \( (1, -5) \)

Evaluate Each Point

1. For point \( (2, 5) \):

  • Check \( y > -2x + 3 \): \[ 5 > -2(2) + 3 \Rightarrow 5 > -4 + 3 \Rightarrow 5 > -1 \quad \text{(True)} \]

  • Check \( y < 3x - 3 \): \[ 5 < 3(2) - 3 \Rightarrow 5 < 6 - 3 \Rightarrow 5 < 3 \quad \text{(False)} \]

2. For point \( (3, 6) \):

  • Check \( y > -2x + 3 \): \[ 6 > -2(3) + 3 \Rightarrow 6 > -6 + 3 \Rightarrow 6 > -3 \quad \text{(True)} \]

  • Check \( y < 3x - 3 \): \[ 6 < 3(3) - 3 \Rightarrow 6 < 9 - 3 \Rightarrow 6 < 6 \quad \text{(False)} \]

3. For point \( (10, 5) \):

  • Check \( y > -2x + 3 \): \[ 5 > -2(10) + 3 \Rightarrow 5 > -20 + 3 \Rightarrow 5 > -17 \quad \text{(True)} \]

  • Check \( y < 3x - 3 \): \[ 5 < 3(10) - 3 \Rightarrow 5 < 30 - 3 \Rightarrow 5 < 27 \quad \text{(True)} \]

4. For point \( (1, -5) \):

  • Check \( y > -2x + 3 \): \[ -5 > -2(1) + 3 \Rightarrow -5 > -2 + 3 \Rightarrow -5 > 1 \quad \text{(False)} \]

Final Conclusion

The point that satisfies both inequalities \( y > -2x + 3 \) and \( y < 3x - 3 \) is:

\( (10, 5) \).

So, the correct response is:

\( (10, 5) \).

Use the image to answer the question.

Four points and two intersecting lines are plotted on a coordinate plane. The x-axis ranges from negative 15 to 5 in unit increments, labeled at every 5 units. The y-axis ranges from negative 5 to 10 in unit increments, labeled at every 5 units.

Which point on the graph is a solution to the simultaneous inequalities y≤13x+2
and y<23x+4
?

(1 point)
Responses

(2,5)
left parenthesis 2 comma 5 right parenthesis

(−10,−2)
left parenthesis negative 10 comma negative 2 right parenthesis

(−6,0)
left parenthesis negative 6 comma 0 right parenthesis

(−4,−2)

To determine which point is a solution for the simultaneous inequalities \( y \leq 13x + 2 \) and \( y < 23x + 4 \), we will evaluate each potential point by substituting the coordinates into both inequalities.

Given Inequalities:

  1. \( y \leq 13x + 2 \)
  2. \( y < 23x + 4 \)

Given Points:

  1. \( (2, 5) \)
  2. \( (-10, -2) \)
  3. \( (-6, 0) \)
  4. \( (-4, -2) \)

Evaluate Each Point:

1. For point \( (2, 5) \):

  • Check \( y \leq 13x + 2 \): \[ 5 \leq 13(2) + 2 \Rightarrow 5 \leq 26 + 2 \Rightarrow 5 \leq 28 \quad \text{(True)} \]

  • Check \( y < 23x + 4 \): \[ 5 < 23(2) + 4 \Rightarrow 5 < 46 + 4 \Rightarrow 5 < 50 \quad \text{(True)} \]

2. For point \( (-10, -2) \):

  • Check \( y \leq 13x + 2 \): \[ -2 \leq 13(-10) + 2 \Rightarrow -2 \leq -130 + 2 \Rightarrow -2 \leq -128 \quad \text{(False)} \]

3. For point \( (-6, 0) \):

  • Check \( y \leq 13x + 2 \): \[ 0 \leq 13(-6) + 2 \Rightarrow 0 \leq -78 + 2 \Rightarrow 0 \leq -76 \quad \text{(False)} \]

4. For point \( (-4, -2) \):

  • Check \( y \leq 13x + 2 \): \[ -2 \leq 13(-4) + 2 \Rightarrow -2 \leq -52 + 2 \Rightarrow -2 \leq -50 \quad \text{(False)} \]

Final Conclusion:

The only point that satisfies both inequalities \( y \leq 13x + 2 \) and \( y < 23x + 4 \) is:

\( (2, 5) \).

So, the correct response is:

\( (2, 5) \).