Linear Inequalities Test -11

Given:

\(a^{2}-b^{2}=88\)

\(a-b=4\)

\(a^{2}-b^{2}=(a-b)(a+b)\)

\(a-b=4\) ......(1)

\((a-b)(a+b)=88\)

\(\Rightarrow 4 \times(a+b)=88\)

\(\Rightarrow a+b=\frac{88}4\)

\(\Rightarrow a+b=22\) ............(2)

Adding equation (1) and equation (2), we get

\(\Rightarrow a-b+a+b=4+22\)

\(2 a=26\)

\(\Rightarrow a=13\)

Put the value of a in equation (2), we get

\(13+b=22\)

\(\Rightarrow b=9\)

value of \(a b=13 \times 9\)

\(\Rightarrow a b=117\)

\(\therefore\) The value of \(a b\) is 117.

Given:

\(-4=-7+3 x\)

\(\Rightarrow 3 x=7-4\)

\(\Rightarrow x=\frac33\)

\(\therefore x=1\)

A triangular region in the 3rd quadrant

Given inequalities x ≥ 0 , y ≥ 0 , 2x + y + 6 ≥ 0

Now take x = 0, y = 0 and 2x + y + 6 = 0

when x = 0, y = -6

when y = 0, x = -3

So, the points are A(0, 0), B(0, -6) and C(-3, 0)

So, the graph of the inequations x ≤ 0 , y ≤ 0 , and 2x + y + 6 ≥ 0 is a triangular region in the 3rd quadrant.

If the system of equations has more than one solution, then \(D=0\) and \(D_{1}=D_{2}=D_{3}=0\). From the given set of equation, we form,

\(D_{1}=0\)

\(\Rightarrow\left|\begin{array}{ccc}a & 2 & 3 \\ b & -1 & 5 \\ c & -3 & 2\end{array}\right|=0\)

Now,

\(\Rightarrow a(-2+15)-2(2 b-5 c)+3(-3 b+c)=0\)

\(\Rightarrow 13 a-4 b+10 c-9 b+3 c=0\)

\(\Rightarrow 13 a-13 b+13 c=0\)

\(\Rightarrow a-b+c=0\)

\(\therefore b-a-c=0\)

Given,

\(\frac{3(x-2)}5 \geq \frac{5(2-x)}3\)

⇒ 3(x – 2) × 3 ≥ 5(2 – x) × 5

⇒ 9(x – 2) ≥ 25(2 – x)

⇒ 9x – 18 ≥ 50 – 25x

⇒ 9x – 18 + 25x ≥ 50

⇒ 34x – 18 ≥ 50

⇒ 34x ≥ 50 + 18

⇒ 34x ≥ 68

\(\Rightarrow x \geq \frac{68}{34}\)

⇒ x ≥ 2

⇒ x ∈ [2, ∞)

Given,

\(\frac{-1}{(|x|-2)}\geq 1\)

⇒\(\frac{-1}{(|x|-2)} \geq 0\)\)

⇒\({\{-1-(|x|-2)\}}{(|x|-2)} \geq 0\)

⇒\({\{1-|x|\}}{(|x|-2)} \geq 0\)

⇒\(\frac{-(|x|-1)}{(|x|-2)} \geq 0\)

Using number line rule:

1 ≤ |x| < 2

⇒ x ∈ (-2, -1) ∪ (1, 2)

Given:

\(9 a-a^{2} \leq 17 a+15\)

On rearranging

\(-a^{2}+9 a \leq 17 a+15\)

Shifting the sign

\(a^{2}-9 a \geq-17 a-15\)

\(a^{2}-9 a+17 a+15 \geq 0\)

\(a^{2}+8 a+15 \geq 0\)

\(a^{2}+5 a+3 a+15 \geq 0\)

\(a(a+5)+3(a+5) \geq 0\)

\((a+3)(a+5) \geq 0\)

So, \(-3\) and \(-5\) are the roots of the equation.

Now look at the diagram given below,

We see that all the numbers less than \(-5\) and all the numbers greater than -3 will give us positive result. While numbers between \(-5\) and \(-3\) will give us negative results.

So, all the above value holds for the above equation.

Given inequalities x ≥ 0, y ≥ 0, 2x + y + 6 ≤ 0

Now take x = 0, y = 0 and 2x + y + 6 = 0

when x = 0, y = -6

when y = 0, x = -3

So, the points are A(0, 0), B(0, -6) and C(-3, 0)

Since region is outside from the line 2x + y + 6 = 0
So, it does not represent any figure.

Given,

\(|\frac2{(x-4)}|>1\)

\(\Rightarrow \frac2{|x-4|}>1\)

\(\Rightarrow 2>|x-4|\)

\(\Rightarrow|x-4|<2\)

\(\Rightarrow-2

\(\Rightarrow-2+4

\(\Rightarrow 2

\(\Rightarrow x \in(2,6)\), where \(x \neq 4\)

\(\Rightarrow x \in(2,4) \cup(4,6)\)

1st inequality:

\(\Rightarrow 5 x-1<3 x+2\)

\(\Rightarrow 2 x<3\)

\(\Rightarrow x<\frac3 2\)

\(\therefore x<1.5\)

2nd inequality:

\(\Rightarrow 5 x+5>6-2 x\)

\(\Rightarrow 7 x>1\)

\(\Rightarrow x>\frac17\)

\(\therefore x>0.142\)

\(\therefore {x}\) ranges between \(0.142\) and \(1.5\)

\(\therefore {x}=1\)