Question

22xplus+33y
equals=
3838
33xplus+22y
equals=
3737

Answer 1

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:

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