22xplus+33y

equals=
3838
33xplus+22y
equals=
3737

1 answer

To solve the system of equations given by:

  1. \( 22x + 33y = 3838 \)
  2. \( 33x + 22y = 3737 \)

We can use the substitution or elimination method. Let's choose the elimination method for this case.

First, let's multiply the first equation by 3 and the second equation by 2 to align the coefficients for \( y \):

  1. \( 3(22x + 33y) = 3(3838) \) \[ 66x + 99y = 11514 \]

  2. \( 2(33x + 22y) = 2(3737) \) \[ 66x + 44y = 7474 \]

Now we have the new system of equations:

  1. \( 66x + 99y = 11514 \)
  2. \( 66x + 44y = 7474 \)

Next, we can eliminate \( x \) by subtracting the second equation from the first:

\[ (66x + 99y) - (66x + 44y) = 11514 - 7474 \]

This simplifies to:

\[ 55y = 4040 \]

Now, solving for \( y \):

\[ y = \frac{4040}{55} = 73.09 \]

Now we can substitute \( y \) back into one of the original equations to find \( x \). Let's use the first original equation:

\[ 22x + 33(73.09) = 3838 \]

Calculating \( 33(73.09) \):

\[ 33 \times 73.09 = 2412.67 \]

Now plug this value into the equation:

\[ 22x + 2412.67 = 3838 \]

Subtract \( 2412.67 \) from both sides:

\[ 22x = 3838 - 2412.67 \]

Calculating \( 3838 - 2412.67 \):

\[ 22x = 1425.33 \]

Now, solving for \( x \):

\[ x = \frac{1425.33}{22} \approx 64.15 \]

Thus, the solution to the system of equations is approximately:

\[ x \approx 64.15, \quad y \approx 73.09. \]