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Q: What is x if x 2y 10 and y 3?

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5x + (3y-x) + (10-2y) = 5x + 3y - x + 10 -2y = 5x - x + 3y - 2y + 10 = 4x + y +10

x + 2y = 10At the x-intercept, y=0:x = 10At the y-intercept, x=0 ==> 2y = 10y = 5

2. If x=5 & y =3 then: 3x + 2y = 3 x 5 + 2 x 3 = 15 + 6 = 21 2x + 3y = 2 x 5 + 3 x 3 = 10 + 9 = 19 ⇒ 3x+2y exceeds 2x+3y by 21 - 19 = 2. Alternatively: 3x+2y exceeds 2x+3y by: (3x+2y) - (2x + 3y) = 3x + 2y - 2x - 3y = x - y If x=5 & y=3 then this becomes: x - y = 5 - 3 = 2

Sum = addition Difference = subtraction Say your first number is x and your second number is y. So: x+y = 10 and x-y = 20 Solve for one variable for one of the above equations and substitute that variable into the other equation: x = 10-y (10-y)-y = 20 10-2y = 20 Get y by itself: -2y = 20-10 -2y = 10 Solve for y: -2y/-2 = 10/-2 y = -5 To solve for x: x = 10-y x = 10-(-5) x = 10+5 x = 15 Therefore, the two numbers you are looking for are -5 and 15.

7x-2y=10 -2y=10-7x y=-5+(7/2)x y=(7/2)x-5 (Final answer)

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x = y+3 x+2y = 9 => x = 9-2y y+3 = 9-2y 2y+y = 9-3 3y = 6 y = 2 and x = 5

10-y=2y+3 10-3=2y+y 7=3y 7/3=y

(x^2 + 2y)(x^3 - 2xy + y^3) = x^2(x^3 - 2xy + y^3) + 2y(x^3 - 2xy + y^3) Now, let's distribute each term: = x^2 * x^3 - x^2 * 2xy + x^2 * y^3 + 2y * x^3 - 2y * 2xy + 2y * y^3 Now, simplify each term: = x^5 - 2x^3y + x^2y^3 + 2x^3y - 4xy^2 + 2y^4 Now, combine like terms: = x^5 + x^2y^3 - 4xy^2 + 2y^4 So, the expanded form of (x^2 + 2y)(x^3 - 2xy + y^3) is x^5 + x^2y^3 - 4xy^2 + 2y^4.

X+2y=6 how to solve? 2y=6-x then y=3-x so y=3-x

3x^2-2y x=3 y=5 3(3)^2-2(5) 3(9)-10 27-10 17 3x^2-2y=17

Any pair of numbers where the 'y' is 5 more than the 'x'. There are an infinite number of suitable pairs. 2y-2x=10 2y-2x+2x=10+2x 2y=2x+10 divide both side by 2. y=x+5

(A) 3x + 2y = -2 (B) 6x - y = 6 (A) + 2(B): 3x + 2y + 12x - 2y = -2 + 12 or 15x = 10 or x = 2/3 substitute this value of x in (A) 2 + 2y = -2 2y = -4 and y = -2 Answer: x = 2/3, y = -2

Assuming y=2x+5 and x=2y-15y=2x+5 (1)x=2y-15 (2)Substitute y=2x+5 into (2):x=2(2x+5)-15x=4x+10-150=3x+10-150=3x-55=3x5/3=x (x=5/3)Substitute x=5/3 into y=2x+5:y=2(5/3)+5y=10/3+5y=25/3x=5/3, y=25/3

5x + (3y-x) + (10-2y) = 5x + 3y - x + 10 -2y = 5x - x + 3y - 2y + 10 = 4x + y +10

x + 2y = 10At the x-intercept, y=0:x = 10At the y-intercept, x=0 ==> 2y = 10y = 5

x + 2y = 10At the x-intercept, y=0:x = 10At the y-intercept, x=0 ==> 2y = 10y = 5

2. If x=5 & y =3 then: 3x + 2y = 3 x 5 + 2 x 3 = 15 + 6 = 21 2x + 3y = 2 x 5 + 3 x 3 = 10 + 9 = 19 ⇒ 3x+2y exceeds 2x+3y by 21 - 19 = 2. Alternatively: 3x+2y exceeds 2x+3y by: (3x+2y) - (2x + 3y) = 3x + 2y - 2x - 3y = x - y If x=5 & y=3 then this becomes: x - y = 5 - 3 = 2