Solve for x and y if x/2+y/4=1/2 and x/3-y/5=1/9
Write x/2+y/4=1/2 as 2x+y=2, thus, y=2-2x:
x/3 - y/5 = 1/9
5x - 3y = 15/9
5x - 3(2-2x) = 15/9
5x - 6 + 6x = 15/9
11x - 6 = 15/9
11x = 69/9
x = 69/99
x = 23/33
y = 2-2x = 2-2(23/33) = 2-(46/33) = 66/33-46/33 = 20/33
Therefore, x=23/33 and y=20/33
Alright, let's solve the system of equations with a touch of humor:
To solve for x and y, we'll use a method called elimination. But it's not the kind of elimination where you send the equations to a deserted island. Instead, we'll multiply them by numbers to get rid of those pesky fractions.
Starting with the first equation x/2 + y/4 = 1/2, let's multiply everything by 4 to say goodbye to those halves. We get:
2x + y = 2
Now, onto the second equation x/3 - y/5 = 1/9. Let's make the numbers play nicely by multiplying everything by 15. This gives us:
5x - 3y = 5/3
Now, let's rearrange the equations and put them side by side:
2x + y = 2
5x - 3y = 5/3
To eliminate y, let's multiply the first equation by 3 and the second equation by 2:
6x + 3y = 6
10x - 6y = 10/3
Now, we'll subtract the first equation from the second equation:
(10x - 6y) - (6x + 3y) = (10/3) - 6
Simplifying this equation skillfully, we get:
10x - 6x - 6y - 3y = 10/3 - 18/3
4x - 9y = -8/3
Hmm, things are getting a bit tricky here. Let's simplify it a bit further by multiplying the entire equation by 3:
12x - 27y = -8
Great! Now we have a new equation. Let's call it equation C. To solve for x, let's think about equation C and the first equation together:
2x + y = 2
12x - 27y = -8
To eliminate y, let's multiply the first equation by 27 and add it to equation C:
(2x + y) + (12x - 27y) = (2) + (-8)
Let's sum things up:
14x - 26y = -6
Ah, this looks simpler already. Let's call this new equation D. Now, let's look at equations C and D together:
12x - 27y = -8
14x - 26y = -6
Hmm, now we have a system of two equations with two variables, x and y. Let's use the elimination method once again. To eliminate x, let's multiply equation C by 14 and equation D by 12:
168x - 378y = -112
168x - 312y = -72
Subtracting equation D from equation C:
(168x - 378y) - (168x - 312y) = -112 - (-72)
Simplifying this skillfully:
-378y + 312y = -112 + 72
-66y = -40
Now, let's divide both sides by -66, but don't forget to swap the signs:
y = 40/66
Simplifying this, we get:
y = 20/33
Now, let's replace y = 20/33 in equation C or D (you can choose). I'll go with equation D for fun:
14x - 26(20/33) = -6
Simplifying this skillfully:
14x - (520/33) = -6
14x = -6 + (520/33)
Combining the terms carefully:
14x = (-198 + 520)/33
Calculating the numerator like a juggling clown:
14x = 322/33
Finally, dividing both sides by 14:
x = (322/33)/14
Reducing this extravagant fraction:
x = 23/33
So, after all the joking, x = 23/33 and y = 20/33. I hope you had as much fun solving the equations as I did explaining them!
To solve for x and y in the given system of equations:
Equation 1: x/2 + y/4 = 1/2
Equation 2: x/3 - y/5 = 1/9
We can start by eliminating the fractions by multiplying both sides of each equation by the least common multiple (LCM) of the denominators, which is 20.
Multiplying equation 1 by 20, we get:
20(x/2) + 20(y/4) = 20(1/2)
10x + 5y = 10
Multiplying equation 2 by 20, we get:
20(x/3) - 20(y/5) = 20(1/9)
20x/3 - 4y = 20/9
Now, we have the following system of equations:
Equation 3: 10x + 5y = 10
Equation 4: 20x/3 - 4y = 20/9
Let's solve this system using the method of substitution.
From Equation 3, we can solve for x in terms of y:
10x = 10 - 5y
x = (10 - 5y)/10
Now, substitute x in Equation 4 with (10 - 5y)/10:
20((10 - 5y)/10)/3 - 4y = 20/9
Simplifying this equation, we get:
(2(10 - 5y)/3) - 4y = 20/9
(20 - 10y)/3 - 4y = 20/9
Next, we can multiply both sides of this equation by 9 to eliminate the fractions:
9(20 - 10y)/3 - 4y * 9 = 20
Expanding and rearranging further, we get:
(180 - 90y)/3 - 36y = 20
(180 - 90y - 108y)/3 = 20
(180 - 198y)/3 = 20
Now, multiply both sides of the equation by 3 to eliminate the denominator:
180 - 198y = 60
Solving for y, we have:
-198y = 60 - 180
-198y = -120
y = (-120)/(-198)
y = 20/33
Now, substitute the value of y back into Equation 3 to solve for x:
10x + 5(20/33) = 10
10x + 100/33 = 10
10x = 10 - 100/33
10x = (330 - 100)/33
10x = 230/33
x = (230/33)/10
x = (230/33) * (1/10)
x = 23/33
Therefore, the solution to the system of equations is:
x = 23/33
y = 20/33
To solve for x and y, we can use the method of substitution. Here's how you can do it step by step:
Step 1: Rearrange the first equation to express x in terms of y. Multiply both sides of the equation by 2 to get rid of the denominator:
x/2 + y/4 = 1/2
Multiply both sides by 2:
x + y/2 = 1
Step 2: Solve the rearranged equation for x:
x = 1 - y/2
Step 3: Substitute the expression for x in terms of y into the second equation:
x/3 - y/5 = 1/9
Substitute x = 1 - y/2:
(1 - y/2)/3 - y/5 = 1/9
Step 4: Simplify the equation:
Multiply through by the common denominator, which is 15:
5(1 - y/2) - 3y = 5/3
Distribute the 5:
5 - 5y/2 - 3y = 5/3
Multiply through by 6 to clear the fractions:
30 - 15y - 18y = 10
Combine like terms:
-33y = -20
Step 5: Solve for y:
Divide both sides by -33:
y = -20/-33
y = 20/33
Step 6: Substitute the value of y back into one of the original equations to solve for x. Let's use the first equation:
x/2 + y/4 = 1/2
Substitute y = 20/33:
x/2 + (20/33)/4 = 1/2
Simplify the equation:
x/2 + 5/33 = 1/2
Subtract 5/33 from both sides:
x/2 = 1/2 - 5/33
Find a common denominator:
x/2 = 33/66 - 10/66
Combine the fractions on the right side:
x/2 = 23/66
Multiply both sides by 2 to solve for x:
x = 2(23/66)
x = 23/33
Therefore, the solution to the system of equations is x = 23/33 and y = 20/33.