y - 2x < 2
-2 - 2x is greater than or equal to 6 only if x is less than or equal to -4.
2 - 2x ≤ 32 - 2 - 2x ≤ 3 - 2-2x ≤ 1-2x/-2 ≥ 1/-2x ≥ -1/2 orx ≥ -0.5-1 ≤ 2x + 1 < 4-1 - 1 ≤ 2x + 1 - 1 < 4 - 1-2 ≤ 2x < 3-2/2 ≤ 2x/2 < 3/2-1 ≤ x < 1.5So the solution set for both inequalities is -0.5 ≤ x < 1.5. The two integers that satisfy both the inequalities are 0 and 1.
2x+3<7 2x<7-3 2x<4 x<2
Since 2x + 3 > 7, then 2x > 4. and x > 2.Of the choices listed, only 0 is not greater than 2.Steps.2x + 3 > 72x + 3 - 3 > 7 - 32x > 42x/2 > 4/2x > 2
2x - 12 > 2 + x x - 12 > 2 x > 14
No. For 0 < x < 2, 2x is larger.
There is no solution to: 4(2x + 2) + 12 > 1004(2x + 2) + 12 > 100 For 4(2x + 2) + 12 > 1004(2x + 2) + 12 to have a solution x < -1 which makes 1004(2x + 2) + 12 < 12, BUT 1004(2x + 2) + 12 is supposed to be greater than 100. Perhaps there is a missing operator in the digits of 1004?
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
2x * 2x (2*2=4) therefore, 2x*2x equals 4x2.
2x - 2
2x+2