9th Class Math Chapter 5 Notes

Q1(i) — Solve: $12x+30=-6$

Solve and represent the solution on a real line.

1. $12x=-36$.

Answer: $x=-3$ (a single point at $-3$ on the number line)

Q1(ii) — Solve: $\frac{x}{3}+6=-12$

Solve and represent the solution on a real line.

1. $\frac{x}{3}=-18$.

Answer: $x=-54$

Q1(iii) — Solve: $\frac{x}{2}-\frac{3x}{4}=\frac{1}{12}$

Solve and represent the solution on a real line.

1. Multiply by LCD $12$: $6x-9x=1\Rightarrow -3x=1$.

Answer: $x=-\frac{1}{3}$

Q1(iv) — Solve: $2=7(2x+4)+12x$

Solve and represent the solution on a real line.

1. $2=14x+28+12x=26x+28$.
2. $-26=26x$.

Answer: $x=-1$

Q1(v) — Solve: $\frac{2x-1}{3}-\frac{3x}{4}=\frac{5}{6}$

Solve and represent the solution on a real line.

1. Multiply by LCD $12$: $4(2x-1)-9x=10\Rightarrow 8x-4-9x=10\Rightarrow -x=14$.

Answer: $x=-14$

Q1(vi) — Solve: $\frac{-5x}{10}=9-\frac{10x}{5}$

Solve and represent the solution on a real line.

1. Simplify: $-\frac{x}{2}=9-2x$.
2. Multiply by $2$: $-x=18-4x\Rightarrow 3x=18$.

Answer: $x=6$

Q2(i) — Solve: $\frac{x+5}{3}=1-x$

Solve and represent the solution on a real line.

1. $x+5=3(1-x)=3-3x\Rightarrow 4x=-2$.

Answer: $x=-\frac{1}{2}$

Q2(ii) — Solve: $\frac{2x+1}{3}+\frac{1}{2}=1-\frac{x-1}{3}$

Solve and represent the solution on a real line.

1. Multiply by LCD $6$: $2(2x+1)+3=6-2(x-1)$.
2. $4x+2+3=6-2x+2\Rightarrow 4x+5=8-2x\Rightarrow 6x=3$.

Answer: $x=\frac{1}{2}$

Q3(i) — Solve: $x-6\le -2$

Solve each inequality and represent the solution on a real line.

1. Add $6$ to both sides.

Answer: $x\le 4$ (ray on the number line, closed at $4$, extending left)

Q3(ii) — Solve: $-9>-16+x$

Solve and represent on a real line.

1. Add $16$: $7>x$, i.e. $x<7$.

Answer: $x<7$ (open ray extending left from $7$)

Q3(iii) — Solve: $3+2x\ge 3$

Solve and represent on a real line.

1. $2x\ge 0$.

Answer: $x\ge 0$ (closed ray extending right from $0$)

Q3(iv) — Solve: $6(x+10)\le 0$

Solve and represent on a real line.

1. Divide by $6$: $x+10\le 0$.

Answer: $x\le -10$

Q3(v) — Solve: $\frac{5}{3}x-\frac{3}{4}<\frac{-1}{12}$

Solve and represent on a real line.

1. Multiply by $12$: $20x-9<-1\Rightarrow 20x<8$.

Answer: $x<\frac{2}{5}$

Q3(vi) — Solve: $\frac{1}{4}x-\frac{1}{2}\le -1+\frac{1}{2}x$

Solve and represent on a real line.

1. Multiply by $4$: $x-2\le -4+2x\Rightarrow -x\le -2$.
2. Divide by $-1$ (flip inequality): $x\ge 2$.

Answer: $x\ge 2$

Q4(i) — Solve: $3x+7<16$

Solve and represent on a real line.

1. $3x<9$.

Answer: $x<3$

Q4(ii) — Solve: $5(x-3)\ge 26x-(10x+4)$

Solve and represent on a real line.

1. $5x-15\ge 26x-10x-4=16x-4$.
2. $5x-16x\ge -4+15\Rightarrow -11x\ge 11$.
3. Divide by $-11$ (flip): $x\le -1$.

Answer: $x\le -1$

Q5(i) — Shade: $2x+y\le 6$

Shade the solution region in the $xy$-plane.

1. Boundary line $2x+y=6$ (intercepts $(3,0)$ and $(0,6)$), drawn solid since $\le$.
2. Test $(0,0)$: $0\le 6$ true — shade the side containing the origin.

Answer: Solid line through $(3,0)$, $(0,6)$; shade the origin's side

Q5(ii) — Shade: $3x+7y\ge 21$

Shade the solution region.

1. Boundary $3x+7y=21$ (intercepts $(7,0)$, $(0,3)$), solid line.
2. Test $(0,0)$: $0\ge 21$ false — shade the side away from the origin.

Answer: Solid line through $(7,0)$, $(0,3)$; shade away from the origin

Q5(iii) — Shade: $3x-2y\ge 6$

Shade the solution region.

1. Boundary $3x-2y=6$ (intercepts $(2,0)$, $(0,-3)$), solid line.
2. Test $(0,0)$: $0\ge 6$ false — shade away from the origin.

Answer: Solid line through $(2,0)$, $(0,-3)$; shade away from the origin

Q5(iv) — Shade: $5x-4y\le 20$

Shade the solution region.

1. Boundary $5x-4y=20$ (intercepts $(4,0)$, $(0,-5)$), solid line.
2. Test $(0,0)$: $0\le 20$ true — shade the origin's side.

Answer: Solid line through $(4,0)$, $(0,-5)$; shade the origin's side

Q5(v) — Shade: $2x+1\ge 0$

Shade the solution region.

1. Simplifies to $x\ge -\frac{1}{2}$: a vertical boundary line.

Answer: Vertical solid line $x=-\frac{1}{2}$; shade everything to the right

Q5(vi) — Shade: $3y-4\le 0$

Shade the solution region.

1. Simplifies to $y\le \frac{4}{3}$: a horizontal boundary line.

Answer: Horizontal solid line $y=\frac{4}{3}$; shade everything below

Q5(vii) — Shade: $2x-3y\le 6$

Shade the solution region.

1. Boundary $2x-3y=6$ (intercepts $(3,0)$, $(0,-2)$), solid line.
2. Test $(0,0)$: $0\le 6$ true — shade the origin's side.

Answer: Solid line through $(3,0)$, $(0,-2)$; shade the origin's side

Q6(i) — Shade: $x+y\ge 5,x-y\le 1$

Indicate the solution region by shading.

1. Line $x+y=5$ (intercepts $(5,0)$,$(0,5)$): test origin fails, shade away from origin.
2. Line $x-y=1$ (intercepts $(1,0)$,$(0,-1)$): test origin holds, shade origin's side.
3. Boundary lines cross at $(3,2)$ (solving simultaneously).

Answer: Overlap region between the two shaded half-planes, meeting at $(3,2)$

Q6(ii) — Shade: $3x+7y\ge 21,x-y\le 2$

Indicate the solution region by shading.

1. Line $3x+7y=21$: shade away from origin.
2. Line $x-y=2$: test origin holds, shade origin's side.
3. Lines cross at $(3.5,1.5)$.

Answer: Overlap region, boundary lines meeting at $(3.5,1.5)$

Q6(iii) — Shade: $4x-3y\le 12,x\ge -\frac{3}{2}$

Indicate the solution region by shading.

1. Line $4x-3y=12$: test origin holds, shade origin's side.
2. Vertical line $x=-\frac{3}{2}$: shade to the right.
3. Lines cross at $\left(-\frac{3}{2},-6\right)$.

Answer: Overlap region, boundary lines meeting at $\left(-\frac{3}{2},-6\right)$

Q6(iv) — Shade: $3x+7y\ge 21,y\le 4$

Indicate the solution region by shading.

1. Line $3x+7y=21$: shade away from origin.
2. Horizontal line $y=4$: shade below.
3. Lines cross at $\left(-\frac{7}{3},4\right)$.

Answer: Overlap region, boundary lines meeting at $\left(-\frac{7}{3},4\right)$

Q6(v) — Shade: $5x+7y\le 35,x-2y\le 2$

Indicate the solution region by shading.

1. Line $5x+7y=35$: test origin holds, shade origin's side.
2. Line $x-2y=2$: test origin holds, shade origin's side.
3. Lines cross at $\left(\frac{84}{17},\frac{25}{17}\right)\approx (4.94,1.47)$.

Answer: Overlap region, boundary lines meeting near $(4.94,1.47)$

Q6(vi) — Solution region: $3x-4y\le 12,3x+2y\ge 3$

Find the solution region.

1. Line $3x-4y=12$: test origin holds, shade origin's side.
2. Line $3x+2y=3$: test origin fails, shade away from origin.
3. Lines cross at $(2,-1.5)$.

Answer: Overlap region, boundary lines meeting at $(2,-1.5)$

Q6(vii) — Solution region: $2x+y\le 4,x+2y\le 6$

Find the solution region.

1. Both lines: test origin holds for each, shade origin's side of both.
2. Lines cross at $\left(\frac{2}{3},\frac{8}{3}\right)$.

Answer: Overlap region, boundary lines meeting at $\left(\frac{2}{3},\frac{8}{3}\right)$

Q7(i) — Maximize $f=2x+5y$ subject to $2y-x\le 8,x-y\le 4,x,y\ge 0$

Maximize $f(x,y)=2x+5y$.

1. Find corner points of the feasible region: $(0,0)$, $(4,0)$ [where $x-y=4$ meets $y=0$], $(0,4)$ [where $2y-x=8$ meets $x=0$], and their intersection $(16,12)$ [solving both lines together].
2. Evaluate $f$: $(0,0)\to 0$; $(4,0)\to 8$; $(16,12)\to 32+60=92$; $(0,4)\to 20$.

Answer: Maximum $f=92$ at $(16,12)$

Q7(ii) — Maximize $f=x+3y$ subject to $2x+5y\le 30,5x+4y\le 20,x,y\ge 0$

Maximize $f(x,y)=x+3y$.

1. Checking both constraints at each axis shows $5x+4y\le 20$ is the binding one throughout — the region is simply the triangle $(0,0),(4,0),(0,5)$.
2. Evaluate $f$: $(0,0)\to 0$; $(4,0)\to 4$; $(0,5)\to 15$.

Answer: Maximum $f=15$ at $(0,5)$

Q7(iii) — Maximize $z=2x+3y$ subject to $2x+y\le 4,4x-y\le 2,x,y\ge 0$

Maximize $z=2x+3y$.

1. Corner points: $(0,0)$, $(0.5,0)$ [where $4x-y=2$ meets $y=0$], $(1,2)$ [intersection of both lines], $(0,4)$ [where $2x+y=4$ meets $x=0$].
2. Evaluate $z$: $(0,0)\to 0$; $(0.5,0)\to 1$; $(1,2)\to 8$; $(0,4)\to 12$.

Answer: Maximum $z=12$ at $(0,4)$

Q7(iv) — Minimize $z=2x+y$ subject to $x+y\ge 3,7x+5y\le 35,x,y\ge 0$

Minimize $z=2x+y$.

1. Corner points: $(3,0)$, $(5,0)$, $(0,7)$, $(0,3)$ (the two lines don't intersect within the first quadrant).
2. Evaluate $z$: $(3,0)\to 6$; $(5,0)\to 10$; $(0,7)\to 7$; $(0,3)\to 3$.

Answer: Minimum $z=3$ at $(0,3)$

Q7(v) — Maximize $f=2x+3y$ subject to $2x+y\le 10,x+2y\le 14,x,y\ge 0$

Maximize $f(x,y)=2x+3y$.

1. Corner points: $(0,0)$, $(5,0)$, $(2,6)$ [intersection], $(0,7)$.
2. Evaluate $f$: $(0,0)\to 0$; $(5,0)\to 10$; $(2,6)\to 22$; $(0,7)\to 21$.

Answer: Maximum $f=22$ at $(2,6)$

Q7(vi) — Min/max of $z=3x+y$ subject to $3x+5y\ge 15,x+3y\le 9,x,y\ge 0$

Find both the minimum and maximum of $z=3x+y$.

1. The two boundary lines meet exactly at $(0,3)$; other corners are $(5,0)$ [where $3x+5y=15$ meets $y=0$] and $(9,0)$ [where $x+3y=9$ meets $y=0$].
2. Evaluate $z$: $(0,3)\to 3$; $(5,0)\to 15$; $(9,0)\to 27$.

Answer: Minimum $z=3$ at $(0,3)$; maximum $z=27$ at $(9,0)$

Q7(vii) — Maximize $g=x+4y$ subject to $x+y\le 4,x,y\ge 0$

Find the maximum value of $g(x,y)=x+4y$.

1. Corner points of this simple triangle: $(0,0)$, $(4,0)$, $(0,4)$.
2. Evaluate $g$: $(0,0)\to 0$; $(4,0)\to 4$; $(0,4)\to 16$.

Answer: Maximum $g=16$ at $(0,4)$

Q7(viii) — Minimize $f=3x+5y$ subject to $x+3y\ge 3,x+y\ge 2,x,y\ge 0$

Find the minimum value of $f(x,y)=3x+5y$.

1. Corner points: $(0,2)$ [where $x+y=2$ meets $x=0$], the lines' intersection $(1.5,0.5)$, and $(3,0)$ [where $x+3y=3$ meets $y=0$].
2. Evaluate $f$: $(0,2)\to 10$; $(1.5,0.5)\to 4.5+2.5=7$; $(3,0)\to 9$.

Answer: Minimum $f=7$ at $(1.5,0.5)$