Please help - I don't know where to begin. My problem is this: Find the numbers "Y" and "Z" if (1 and 3/5 times "Y") + (3/5 times "Z") = 5 AND (1 and 3/5 times "Z") + (3/5 times "Y") = 6. Thank You!
change 1 and 3/5 to another fraction:
8/5
so the equation is (8/5)Z+ (3/5)Z=5
now you can add 8/5 and 3/5 together because they both share Z.
so it would be 11/5Z=5. Multiply both sides by 5 to get 11Z=5 and divide 5 to both sides to get Z.
the second equation:
8/5(Z)+(3/5)Y=6
since you already know Z, plug it in, and find Y
Thank you!
To solve the first equation, let's rewrite it:
(8/5)Z + (3/5)Z = 5
Combining the terms with Z, we have:
(11/5)Z = 5
To solve for Z, we need to isolate it. Multiply both sides of the equation by the reciprocal of (11/5), which is (5/11):
Z = (5/11) * 5
Z = 25/11
Therefore, Z is equal to 25/11.
Now let's use this value of Z to solve the second equation:
(8/5)Z + (3/5)Y = 6
Substituting Z = 25/11:
(8/5)(25/11) + (3/5)Y = 6
Multiplying the fractions, we get:
(200/55) + (3/5)Y = 6
Simplifying the first term:
(40/11) + (3/5)Y = 6
To isolate Y, we need to move the constant term to the other side:
(3/5)Y = 6 - (40/11)
Find a common denominator of 5 and 11, which is 55:
(3/5)Y = (330/55) - (200/55)
(3/5)Y = 130/55
To simplify the fractions on the right side:
(3/5)Y = (26/11)
To solve for Y, multiply both sides by the reciprocal of (3/5), which is (5/3):
Y = (5/3) * (26/11)
Y = 10/3
Therefore, Y is equal to 10/3.
So the numbers Y and Z that satisfy both equations are Y = 10/3 and Z = 25/11.