StudentAid
HomeLearn
9th Class10th Class11th Class12th ClassPast papers & testsBrowse everything
For schoolsFor driversFor parentsBook a demo
Learn/ 9th Mathematics/ Unit 5 /Numericals

Unit 5: Linear Equations and Inequalities — Exercise 5 1

9th Class Mathematics · Unit 5: Linear Equations and Inequalities

5.1.1.Solve and represent the solution on a real line. (i) 12x + 30 = -6 (ii) x/3 + 6 = -12 (iii) x/2 - 3x/4 = 1/12 (iv) 2 = 7(2x+4) + 12x (v) (2x-1)/3 - 3x/4 = 5/6 (vi) -5x/10 = 9 - 10x/5
Given
(i) 12x + 30 = -6
(i) 12x = -6 - 30 = -36
Result
(i) x = -3
Given
(ii) x3 + 6 = -12
(ii) x3 = -18 ⇒ x = -18 × 3
Result
(ii) x = -54
Given
(iii) x2 - 3x4 = 112
(iii) 12 × x2 - 12 × 3x4 = 12 × 112 ⇒ 6x - 9x = 1 ⇒ -3x = 1
Result
(iii) x = -13
Given
(iv) 2 = 7(2x+4) + 12x
(iv) 2 = 14x + 28 + 12x ⇒ 2 - 28 = 26x ⇒ -26 = 26x
Result
(iv) x = -1
Given
(v) 2x-13 - 3x4 = 56
(v) 12 × 2x-13 - 12 × 3x4 = 12 × 56 ⇒ 4(2x-1) - 9x = 10 ⇒ 8x - 4 - 9x = 10 ⇒ -x = 14
Result
(v) x = -14
Given
(vi) -5x10 = 9 - 10x5
(vi) 10 × left(-5x10right) = 10 × 9 - 10 × left(10x5right) ⇒ -5x = 90 - 20x ⇒ 15x = 90
Result
(vi) x = 6
5.1.2.Solve each inequality and represent the solution on a real line. (i) x - 6 ≤ -2 (ii) -9 > -16 + x (iii) 3 + 2x ≥ 3 (iv) 6(x+10) ≤ 0 (v) (5/3)x - 3/4 < -1/12 (vi) (1/4)x - 1/2 ≤ -1 + (1/2)x
Given
(i) x - 6 le -2
Result
(i) x le 4
Given
(ii) -9 > -16 + x
(ii) -9+16 > x ⇒ 7 > x
Result
(ii) x < 7
Given
(iii) 3 + 2x ge 3
(iii) 2x ge 0
Result
(iii) x ge 0
Given
(iv) 6(x+10) le 0
(iv) 6x + 60 le 0 ⇒ 6x le -60
Result
(iv) x le -10
Given
(v) 53x - 34 < -112
(v) 4(5x) - 9 < -1 ⇒ 20x < 8
Result
(v) x < 25
Given
(vi) 14x - 12 le -1 + 12x
(vi) x - 2 le -4 + 2x ⇒ -2+4 le 2x-x
Result
(vi) x ge 2
5.1.3.Shade the solution region for the following linear inequalities in xy-plane: (i) 2x + y ≤ 6 (ii) 3x + 7y ≥ 21 (iii) 3x - 2y ≥ 6 (iv) 5x - 4y ≤ 20 (v) 2x + 1 ≥ 0 (vi) 3y - 4 ≤ 0
Given
(i) 2x+yle 6
Working
(i) Associated eq. 2x+y=6; x=0 ⇒ (0,6); y=0 ⇒ (3,0)
Result
(i) (0,0):0<6 true, region towards origin
Given
(ii) 3x+7yge 21
Working
(ii) Associated eq. 3x+7y=21; x=0 ⇒ (0,3); y=0 ⇒ (7,0)
Result
(ii) (0,0):0>21 false, region away from origin
Given
(iii) 3x-2yge 6
Working
(iii) Associated eq. 3x-2y=6; x=0 ⇒ (0,-3); y=0 ⇒ (2,0)
Result
(iii) (0,0):0>6 false, region away from origin
Given
(iv) 5x-4yle 20
Working
(iv) Associated eq. 5x-4y=20; x=0 ⇒ (0,-5); y=0 ⇒ (4,0)
Result
(iv) (0,0):0<20 true, region towards origin
Given
(v) 2x+1ge 0
Working
(v) Associated eq. 2x+1=0 ⇒ x=-12
Result
(v) x=0: 1>0 true, region towards origin
Given
(vi) 3y-4le 0
Working
(vi) Associated eq. 3y-4=0 ⇒ y=43
Result
(vi) y=0: 0<4 true, region towards origin
5.1.4.Indicate the solution region of the following linear inequalities by shading: (i) 2x-3y≤6, 2x+3y≤12 (ii) x+y≥5, -y+x≤1 (iii) 3x+7y≥21, x-y≤2 (iv) 4x-3y≤12, x≥-3/2 (v) 3x+7y≥21, y≤4 (vi) 5x+7y≤35, x-2y≤2
Given
(i) 2x-3yle 6;quad 2x+3yle 12
Working
(i) 2x-3y=6: (0,-2),(3,0);quad 2x+3y=12: (0,4),(6,0)
Result
(i) Both 0<6, 0<12 true, region towards origin for both
Given
(ii) x+yge 5;quad x-yle 1
Working
(ii) x+y=5: (0,5),(5,0);quad x-y=1: (0,-1),(1,0)
Result
(ii) 0>5 false (away); 0<1 true (towards)
Given
(iii) 3x+7yge 21;quad x-yle 2
Working
(iii) 3x+7y=21: (0,3),(7,0);quad x-y=2: (0,-2),(2,0)
Result
(iii) 0>21 false (away); 0<2 true (towards)
Given
(iv) 4x-3yle 12;quad xge -32
Working
(iv) 4x-3y=12: (0,-4),(3,0);quad x=-32
Result
(iv) 0<12 true (towards); 0>-32 true (towards)
Given
(v) 3x+7yge 21;quad yle 4
Working
(v) 3x+7y=21: (0,3),(7,0);quad y=4
The book mislabels the associated equation as '3x+7y=12' in the text line, but then correctly uses the intercepts (0,3) and (7,0) (which belong to 3x+7y=21, matching the given stem) and labels the graphed line '3x+7y=21'. The '12' is a typo.
Result
(v) 0>21 false (away); 0<4 true (towards)
Given
(vi) 5x+7yle 35;quad x-2yle 2
Working
(vi) 5x+7y=35: (0,5),(7,0);quad x-2y=2: (0,-1),(2,0)
Result
(vi) 0<35 true (towards); 0<2 true (towards)