3 (x - y)
If x+y = 3 and y times x = 2 then y = 1.
Then y = 3x.
If y varies directly as x and y is 36 when x is 9, then y is always four times the value of x. So if y is 12, then x is 3.
Let one of the numbers be x and other be y.Sum is 33: x + y = 33 ...(1)Difference is 3: x - y = 3 ...(2)Let's add these equations:(x + y) + (x - y) = 33 + 32x = 36x = 18Putting the value of x in equation 1, we get18 + y = 33y = 33 - 18 = 15The two numbers are 18 and 15.15 and 18
3 (x - y)
x5 You have x*x*x*x*x just add up the amount of x's you have and put it after as an exponent Example: 3*3*3*3=34 y*y*y*y*y*y=y6
3[(x^2) + 5] x (2x-y) Do the parentheses first, then brackets next. P.e.m.d.a.s.
If you meant (45x^4)(y), then the answer is: (3)(3)(5)(x)(x)(x)(x)(y) If you meant 45 times x times 4 times y, then the answer is: (2)(2)(3)(3)(5)(x)(y)
(9) · (x) · (2/3) · (y) = 6 x y
-3X - 3Y x 8
That verbal expression would read: 3(x+y)^2 + 5(2x -y) ^2 means 'squared' in the above example.
Two integers (X & Y). X+Y=-3, X-Y=-11. x=-11+y --> (x)+y=-3 --> (-11+y)+y=-3 --> y=4 x=-11+4=-7 Hope that helps! Two integers (X & Y). X+Y=-3, X-Y=-11. x=-11+y --> (x)+y=-3 --> (-11+y)+y=-3 --> y=4 x=-11+4=-7 Hope that helps!
If x+y = 3 and y times x = 2 then y = 1.
2 times 2 times 3 times 3 times x times x times y
x + y = 11 and x - y = 5. The y's cancel so you're left with 2x = 16. So x = 8. Plug x back in to one of the original equations to find y. 8 + y = 11. So y = 3. And 3 times 8 = 24.
Let the two numbers be x and y, then:Twice the first equals three times the second: 2x = 3yThree times their difference exceeds the second by 13: 3(x - y) = y + 13From equation (1) it is clear the first number (x) is greater than the second (y), so their difference is x - y.Equation (2) can be rearranged:3(x - y) = y + 13→ 3x - 3y = y + 13→ 3x = 4y + 13This gives two simultaneous equations:2x - 3y = 03x - 4y = 13which can then be solved to find the two numbers (x and y).