Answer :
Certainly! Let's solve the system of equations step by step:
The given system of equations is:
1) [tex]\(\frac{x}{4} + \frac{y}{5} = 22\)[/tex]
2) [tex]\(\frac{x}{5} + \frac{y}{4} - 1 = 22\)[/tex]
First, we simplify the second equation:
[tex]\(\frac{x}{5} + \frac{y}{4} - 1 = 22\)[/tex]
Adding 1 to both sides, we get:
[tex]\(\frac{x}{5} + \frac{y}{4} = 23\)[/tex]
Now, we have the following system of equations:
1) [tex]\(\frac{x}{4} + \frac{y}{5} = 22\)[/tex]
2) [tex]\(\frac{x}{5} + \frac{y}{4} = 23\)[/tex]
Let's rewrite these equations in a form that eliminates the fractions by multiplying through by the common denominators.
For the first equation:
Multiply everything by 20 (the least common multiple of 4 and 5):
[tex]\(20 \cdot \left(\frac{x}{4}\right) + 20 \cdot \left(\frac{y}{5}\right) = 20 \cdot 22\)[/tex]
This simplifies to:
[tex]\(5x + 4y = 440\)[/tex] ...(3)
For the second equation:
Multiply everything by 20 (the least common multiple of 5 and 4):
[tex]\(20 \cdot \left(\frac{x}{5}\right) + 20 \cdot \left(\frac{y}{4}\right) = 20 \cdot 23\)[/tex]
This simplifies to:
[tex]\(4x + 5y = 460\)[/tex] ...(4)
So now we have a new system of linear equations:
3) [tex]\(5x + 4y = 440\)[/tex]
4) [tex]\(4x + 5y = 460\)[/tex]
Next, we solve this system using the elimination method.
Multiply equation (3) by 4:
[tex]\(20x + 16y = 1760\)[/tex] ...(5)
Multiply equation (4) by 5:
[tex]\(20x + 25y = 2300\)[/tex] ...(6)
Now, subtract equation (5) from equation (6):
[tex]\((20x + 25y) - (20x + 16y) = 2300 - 1760\)[/tex]
[tex]\(20x + 25y - 20x - 16y = 540\)[/tex]
[tex]\(9y = 540\)[/tex]
Divide by 9:
[tex]\(y = 60\)[/tex]
Now substitute [tex]\(y = 60\)[/tex] back into equation (3) to find [tex]\(x\)[/tex]:
[tex]\(5x + 4(60) = 440\)[/tex]
[tex]\(5x + 240 = 440\)[/tex]
Subtract 240 from both sides:
[tex]\(5x = 200\)[/tex]
Divide by 5:
[tex]\(x = 40\)[/tex]
Therefore, the solution to the system of equations is:
[tex]\(x = 40\)[/tex] and [tex]\(y = 60\)[/tex].
The given system of equations is:
1) [tex]\(\frac{x}{4} + \frac{y}{5} = 22\)[/tex]
2) [tex]\(\frac{x}{5} + \frac{y}{4} - 1 = 22\)[/tex]
First, we simplify the second equation:
[tex]\(\frac{x}{5} + \frac{y}{4} - 1 = 22\)[/tex]
Adding 1 to both sides, we get:
[tex]\(\frac{x}{5} + \frac{y}{4} = 23\)[/tex]
Now, we have the following system of equations:
1) [tex]\(\frac{x}{4} + \frac{y}{5} = 22\)[/tex]
2) [tex]\(\frac{x}{5} + \frac{y}{4} = 23\)[/tex]
Let's rewrite these equations in a form that eliminates the fractions by multiplying through by the common denominators.
For the first equation:
Multiply everything by 20 (the least common multiple of 4 and 5):
[tex]\(20 \cdot \left(\frac{x}{4}\right) + 20 \cdot \left(\frac{y}{5}\right) = 20 \cdot 22\)[/tex]
This simplifies to:
[tex]\(5x + 4y = 440\)[/tex] ...(3)
For the second equation:
Multiply everything by 20 (the least common multiple of 5 and 4):
[tex]\(20 \cdot \left(\frac{x}{5}\right) + 20 \cdot \left(\frac{y}{4}\right) = 20 \cdot 23\)[/tex]
This simplifies to:
[tex]\(4x + 5y = 460\)[/tex] ...(4)
So now we have a new system of linear equations:
3) [tex]\(5x + 4y = 440\)[/tex]
4) [tex]\(4x + 5y = 460\)[/tex]
Next, we solve this system using the elimination method.
Multiply equation (3) by 4:
[tex]\(20x + 16y = 1760\)[/tex] ...(5)
Multiply equation (4) by 5:
[tex]\(20x + 25y = 2300\)[/tex] ...(6)
Now, subtract equation (5) from equation (6):
[tex]\((20x + 25y) - (20x + 16y) = 2300 - 1760\)[/tex]
[tex]\(20x + 25y - 20x - 16y = 540\)[/tex]
[tex]\(9y = 540\)[/tex]
Divide by 9:
[tex]\(y = 60\)[/tex]
Now substitute [tex]\(y = 60\)[/tex] back into equation (3) to find [tex]\(x\)[/tex]:
[tex]\(5x + 4(60) = 440\)[/tex]
[tex]\(5x + 240 = 440\)[/tex]
Subtract 240 from both sides:
[tex]\(5x = 200\)[/tex]
Divide by 5:
[tex]\(x = 40\)[/tex]
Therefore, the solution to the system of equations is:
[tex]\(x = 40\)[/tex] and [tex]\(y = 60\)[/tex].
