<p><strong>51. x + 2y ≤ 10, x + y ≥ 1, x – y ≤ 0, x ≥ 0, y ≥ 0</strong></p>

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51. x + 2y ≤ 10, x + y ≥ 1, x – y ≤ 0, x ≥ 0, y ≥ 0

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Payal Gupta | Contributor-Level 10

4 months ago

51. The given system of inequality is

x+2y≤ 10- (1)

x+y≥ 1 - (2)

x – y ≤ 0 - (3)

x≥ 0 and y≥ 0 - (4)

The corresponding equation of (1), (2) and (3) are

x + 2y = 10

x	0	10
y	5	0

and x + y =1

x	0	1
y	1	0

and x – y = 0

x	0	1
y	0	1

Putting (2,0)= (x, y) in inequality (1), (2) and (3),

2+2 × 0 ≤ 10 => 2≤ 10 is true.

and 2+0 ≥ 1 => 2 ≥ 1 is true.

and 2 – 0 ≤ 0 => 2 ≤ 0 is false.

So, the solution of inequality (1) and (2) is the plane that includes point (2,0) whereas the solution of inequality (3) is the plane which includes point (2, 0)

∴ The shaded region represents the solution of the given system of inequal

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(ii) If $\frac{3}{4} - x \leq - 3,$ then $x \dots 4$ .

(iii) If $\frac{2}{x + 2} > 0,$ then $x \dots - 2$ .

(iv) If $x > - 5$ , then $4 x \dots - 20$ .

(v) If $x > y$ and $z < 0$ , then $- x z \dots - y z$ .

(vi) If $p > 0$ and $q < 0$ , then $p - q \dots p$ .

(vii) If $x + 2 > 5$ , then $x \dots - 7$ or $x \dots 3$ .

(viii)If $- 2 x + 1 \geq 9$ , then $x \dots - 4$ .

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This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

$\begin{array}{l} (i) - 4 x \geq 12 \Rightarrow x \leq - 3 \\ H e n c e, t h e f i l l e r i s (\leq) \\ (i i) I f - \frac{3}{4} x \leq - 3 \Rightarrow x \geq 3 x \times \frac{4}{3} \Rightarrow x \geq 4 \\ H e n c e, t h e f i l l e r i s (\geq) \\ (i i i) I f \frac{2}{x + 2} > 0 \Rightarrow x > - 2 \\ H e n c e, t h e f i l l e r i s (>) \\ (i v) I f x > - 5 \Rightarrow 4 x > - 20 \\ H e n c e, t h e f i l l e r i s (>) \\ (v) I f x > y a n d z < 0, t h e n \\ x z < y z \Rightarrow - x z > - y z \\ H e n c e, t h e f i l l e r i s (>) \\ (v i) I f p > 0 a n d q < 0, t h e n \\ p - q > p \\ H e n c e, t h e f i l l e r i s (>) \\ (v i i) I f | x + 2 | > 5 t h e n \\ x + 2 < - 5 o r x + 2 > 5 \\ \Rightarrow x < - 5 - 2 o r x > 5 - 2 \\ \Rightarrow x < - 7 o r x > 3 \\ S o, x \in (- \infty, - 7) \cup (3, \infty) \\ H e n c e, t h e f i l l e r i s (<) o r (>) \\ (v i i i) I f - 2 x + 1 \geq 9 t h e n \\ \Rightarrow - 2 x \geq 9 - 1 \Rightarrow - 2 x \geq 8 \Rightarrow 2 x \leq - 8 \Rightarrow x \leq - 4 \\ H e n c e, t h e f i l l e r i s (\leq) \end{array}$

(i) If $x < y$ and $b < 0$ , then $\frac{x}{b} > \frac{y}{b} .$

(ii) If $x y > 0$ , then $x > 0$ and $y < 0$ .

(iii) If $x y > 0$ , then $x < 0$ and $y < 0$ .

(iv) If $x y < 0$ , then $x < 0$ and $y < 0$ .

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(vi) If $x < - 5$ and $x > 2$ , then $x \in (- 5, 2)$ .

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(x) Graph of $x < 3$ corresponds to Fig 6.12.

(xi) Graph of $x \geq 0$ corresponds to Fig 6.13.

(xii) Graph of $y \leq 0$ corresponds to Fig 6.14.

(xiii) Solution set of $x \geq 0$ and $y \leq 0$ corresponds to Fig 6.15.

(xiv) Solution set of $x \geq 0$ and $y \leq 1$ corresponds to Fig 6.16.

(xv) Solution set of $x + y \geq 0$ corresponds to Fig 6.17.

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51. x + 2y ≤ 10, x + y ≥ 1, x – y ≤ 0, x ≥ 0, y ≥ 0

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