3x + 12 = 42 3x = 30 x = 10
2x+3x = 5x
If: 3x = 15 Then: x = 5
y = 3x 3x + 4y = 30 3x + 4(3x) = 30 3x + 12x = 30 15x = 30 x = 2 y = 3x y = 3(2) y = 6 (2, 6)
3x+7=37 -7 -7 3x=30 /3 /3 x=10
3x+42=6x+12 42=6x-3x+12 42=3x+12 42-12=3x 30=3x 10=x
3x=(33+42) 3x=75 x=75/3 x=25
(42/3x)*12=24 42/3X=2 42=6X x=7 if it'isn't 3x but x^3 then 21=X^3 21^(1/3)=2.76
42
All that can be deduced is that x + y = 14.
3x + 9 = 42 3x = 33 x = 11
42
3x + 12 = 42 3x = 30 x = 10
T = 3x + 7y; x = 4; y = 6T = 3x + 7y (substitute 4 for x, and 6 for y)T = 3(4) + 7(6) = 12 + 42 = 54
3x-1=11 solves to x=4. Plug in 4 to get 42+4. The answer is 20.
3x+2y = 0 -9x+y = 42 => y = 42+9x Substitute the bottom rearranged equation into the top equation: 3x+2(42+9x) = 0 3x+82+18x = 0 21x = -84 Divide both sides by 21 to find the value of x: x = -4 Substitute the value of x into the original equations to find the value of y: So: x = -4 and y = 6
If 15 + 3x = 42, then x = 9