Review the graphs of a system of two linear equations in two variables: x−5y=7

and 3x−2y=−4. Find the solution to both equations.

(1 point)

11 answers

To find the solution to the system of two linear equations

  1. \( x - 5y = 7 \)
  2. \( 3x - 2y = -4 \)

we can solve the equations using the substitution or elimination method. Here, I will use the substitution method.

Step 1: Solve the first equation for \( x \).

From the first equation:

\[ x - 5y = 7 \]

We can rearrange it to solve for \( x \):

\[ x = 5y + 7 \]

Step 2: Substitute \( x \) in the second equation.

Now substitute \( x \) in the second equation \( 3x - 2y = -4 \):

\[ 3(5y + 7) - 2y = -4 \]

Step 3: Simplify and solve for \( y \).

Distributing the 3:

\[ 15y + 21 - 2y = -4 \]

Combine like terms:

\[ 13y + 21 = -4 \]

Next, subtract 21 from both sides:

\[ 13y = -4 - 21 \]

\[ 13y = -25 \]

Now, divide by 13:

\[ y = -\frac{25}{13} \]

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

Now plug the value of \( y \) back into the expression for \( x \):

\[ x = 5\left(-\frac{25}{13}\right) + 7 \]

\[ x = -\frac{125}{13} + 7 \]

Convert 7 to have a common denominator:

\[ x = -\frac{125}{13} + \frac{91}{13} \]

Combine the fractions:

\[ x = \frac{-125 + 91}{13} = \frac{-34}{13} \]

Conclusion:

The solution to the system of equations is:

\[ \left( x, y \right) = \left( -\frac{34}{13}, -\frac{25}{13} \right) \]

Thus, the final answer is:

\[ \boxed{\left(-\frac{34}{13}, -\frac{25}{13}\right)} \]

what is the intersection point to the question above?

The intersection point of the two linear equations, which corresponds to the solution we found earlier, is given by the coordinates \( (x, y) \).

From the previous calculations, we found:

\[ x = -\frac{34}{13} \] \[ y = -\frac{25}{13} \]

So, the intersection point of the two lines given by the equations \( x - 5y = 7 \) and \( 3x - 2y = -4 \) is:

\[ \left( -\frac{34}{13}, -\frac{25}{13} \right) \]

Thus, the intersection point is:

\[ \boxed{\left( -\frac{34}{13}, -\frac{25}{13} \right)} \]

Review the graphs of a system of two linear equations in two variables: 4x+3y=15
and y=x+5
. Find the solution to both equations.

(1 point)
The intersection point is (
).

To find the solution to the system of equations

  1. \( 4x + 3y = 15 \)
  2. \( y = x + 5 \)

we can use substitution or elimination. Here, we'll use substitution since the second equation is already solved for \( y \).

Step 1: Substitute \( y \) in the first equation.

From the second equation, we know:

\[ y = x + 5 \]

Now substitute \( y \) into the first equation:

\[ 4x + 3(x + 5) = 15 \]

Step 2: Simplify and solve for \( x \).

Distributing the 3 gives us:

\[ 4x + 3x + 15 = 15 \]

Combine like terms:

\[ 7x + 15 = 15 \]

Next, subtract 15 from both sides:

\[ 7x = 0 \]

Now divide by 7:

\[ x = 0 \]

Step 3: Substitute \( x \) back to find \( y \).

Now plug the value of \( x \) back into the second equation:

\[ y = 0 + 5 \]

\[ y = 5 \]

Conclusion

The intersection point of the two equations is:

\[ \left( x, y \right) = (0, 5) \]

Thus, the intersection point is:

\[ \boxed{(0, 5)} \]

Review the graphs of a system of two linear equations in two variables: y=x−6
and y=−x+4
. Find the solution to both equations.

(1 point)

To find the solution to the system of equations

  1. \( y = x - 6 \)
  2. \( y = -x + 4 \)

we will set the two expressions for \( y \) equal to each other since both equations equal \( y \):

Step 1: Set the equations equal to each other.

\[ x - 6 = -x + 4 \]

Step 2: Solve for \( x \).

Add \( x \) to both sides:

\[ x + x - 6 = 4 \]

This simplifies to:

\[ 2x - 6 = 4 \]

Next, add 6 to both sides:

\[ 2x = 10 \]

Now, divide by 2:

\[ x = 5 \]

Step 3: Substitute \( x \) back to find \( y \).

Now, substitute \( x = 5 \) back into either equation. We'll use the first equation:

\[ y = 5 - 6 \]

Thus,

\[ y = -1 \]

Conclusion

The solution to the system of equations is:

\[ (x, y) = (5, -1) \]

Thus, the intersection point is:

\[ \boxed{(5, -1)} \]

Estimate the intersection point of the graphs of the following linear equations in two variables.
y=x+2


y=−x+6


(1 point)

To find the intersection point of the two linear equations:

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

we can set the two equations equal to each other since both are equal to \( y \):

Step 1: Set the equations equal to each other.

\[ x + 2 = -x + 6 \]

Step 2: Solve for \( x \).

Add \( x \) to both sides:

\[ x + x + 2 = 6 \]

This simplifies to:

\[ 2x + 2 = 6 \]

Now, subtract 2 from both sides:

\[ 2x = 4 \]

Then divide by 2:

\[ x = 2 \]

Step 3: Substitute \( x \) back to find \( y \).

Now substitute \( x = 2 \) back into either equation. Let's use the first equation:

\[ y = 2 + 2 \]

Thus,

\[ y = 4 \]

Conclusion

The intersection point of the two equations is:

\[ (x, y) = (2, 4) \]

Therefore, the estimated intersection point is:

\[ \boxed{(2, 4)} \]

Estimate the intersection point of the graphs of the following linear equations in two variables.

y=−x+7


y=4x−3


(1 point)

To find the intersection point of the two linear equations:

  1. \( y = -x + 7 \)
  2. \( y = 4x - 3 \)

we set the two equations equal to each other since both are equal to \( y \):

Step 1: Set the equations equal to each other.

\[ -x + 7 = 4x - 3 \]

Step 2: Solve for \( x \).

First, add \( x \) to both sides:

\[ 7 = 5x - 3 \]

Now, add 3 to both sides:

\[ 10 = 5x \]

Next, divide by 5:

\[ x = 2 \]

Step 3: Substitute \( x \) back to find \( y \).

Now substitute \( x = 2 \) back into either equation. We'll use the first equation:

\[ y = -2 + 7 \]

Thus,

\[ y = 5 \]

Conclusion

The intersection point of the two equations is:

\[ (x, y) = (2, 5) \]

Therefore, the estimated intersection point is:

\[ \boxed{(2, 5)} \]