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Solve : \( – 4 \le \left {7 – x} \right < 5\), for x is
If \(9\sqrt x = \sqrt {12} + \sqrt {147} \), then the value of x is
Solve:\( – 3x + 19 \ge 0,3 – 5x < 0\)
Solve the system of inequalities \( – 2 \le 6x – 1 < 2\)
Solve the system of inequalities\( – 15 < \dfrac{{3\left( {x – 2} \right)}}{5} \le 0\)
The solution set for \(\left x \right > 9\) is
The solution set for : \(\left {\dfrac{{2(5 – x)}}{3} < \dfrac{5}{9}} \right\)
Solve the system of inequalities \(\left( {x + 5} \right) – 7\left( {x – 2} \right) \ge 4x + 9, 2\left( {x + 3} \right) – 7\left( {x + 5} \right) \le 3x – 9\)
i.e
If \( – 2x + 1 \ge 9\), then x is
If \(\left {x + 2} \right > 5\), then
or
or
If \(\dfrac{2}{{x – 9}} > 0\) then
when
If \(\dfrac{5}{{\left {11 – x} \right}} > 0\), then x is
when
Solve: \(\dfrac{9}{{x + 3}} \le 4 \le \dfrac{{11}}{{x + 2}}\) for x
\(\dfrac{1}{{\left {x – 3} \right}} \le \dfrac{2}{3}\), Solve for x
but
Solve: \( – 7 \le \dfrac{{3x – 4}}{7} \le 11\), for x
For any a,b,c and d, if \(a + b \le c + d\) and \(a \cdot c \le c \cdot d\) then
As a,b,c,d any real mumbers, the
above inequalities based on their various value.
So option 1,2,3 are not the parcticular solution.
If\(\dfrac{{\left {x – 3} \right}}{9} \le 4\), then x is
Solve: \(7x – 9 \ge 4x + 15,3x – 7 \le 5\)
If \(\left {x – 9} \right \le 8\), then
If \(\dfrac{1}{{a – bx}} < 3\), then x is not equal to
, where
Solve: \({\left( {x + 1} \right)^2} + {({x^2} + 3x + 2)^2} = 0\)
so,
Now, the common solution is x=1
x=1 is the solution of this above equation.
If \(p>q\) and \(q<0\), then pq is
If p < q and q > 0, then pq is
for
So, the value of pq depends on p.
\(\dfrac{2}{{x + 2}} \le 5 \le \dfrac{9}{{x + 1}}\), solve for x
from (i) & (ii)
Solve: \( – 4x + 11 \ge 7\)
Solve: \( – 4 \le \left {2x – 3} \right\), for x is
If \(\dfrac{{ – 3}}{4}x \le – 3\), then x is
Solve: \(\dfrac{{3 – \left x \right}}{4} \le 0\)