Graphical Solution (Simultaneous Linear Equations, Graphically) — ICSE Class 9 Maths Important Questions

Answer: (b) The point where the two lines intersect.

The intersection point has coordinates $(x,y)$(x,y) that satisfy both equations at once, which is exactly the common solution.

Answer: (a) Parallel to the $x$x-axis.

$y=3$y=3 means the ordinate is $3$3 for every $x$x; all such points lie on a horizontal line three units above the $x$x-axis.

Answer: (c) No solution.

Parallel distinct lines never meet, so there is no common point and hence no solution; the system is inconsistent.

Answer: (b) $(3,1)$(3,1).

Adding: $2x=6\Rightarrow x=3$2x=6 x=3; then $3+y=4\Rightarrow y=1$3+y=4 y=1. Both lines pass through $(3,1)$(3,1), so that is the intersection.

Answer: (a) Dividing the second equation by $2$2 gives $x+y=5$x+y=5, the same line as the first. Coincident lines share all points, so there are infinitely many solutions; R correctly explains A.

Substitute each $x$x:

$x=0:\ y=2(0)-1=-1$x=0: y=2(0)-1=-1.
$x=1:\ y=2(1)-1=1$x=1: y=2(1)-1=1.
$x=2:\ y=2(2)-1=3$x=2: y=2(2)-1=3.

The points are $(0,-1),\ (1,1),\ (2,3)$(0,-1), (1,1), (2,3). Plotting and joining these gives the straight line $y=2x-1$y=2x-1.

$x$x-axis: put $y=0$y=0: $2x=12\Rightarrow x=6$2x=12 x=6, giving $(6,0)$(6,0).

$y$y-axis: put $x=0$x=0: $3y=12\Rightarrow y=4$3y=12 y=4, giving $(0,4)$(0,4).

So the line meets the axes at $(6,0)$(6,0) and $(0,4)$(0,4); these two intercepts are enough to draw it.

For $x+y=6$x+y=6 (i.e. $y=6-x$y=6-x):
$x=0\Rightarrow y=6$x=0 y=6; $x=2\Rightarrow y=4$x=2 y=4; $x=6\Rightarrow y=0$x=6 y=0. Points: $(0,6),(2,4),(6,0)$(0,6),(2,4),(6,0).

For $x-y=2$x-y=2 (i.e. $y=x-2$y=x-2):
$x=0\Rightarrow y=-2$x=0 y=-2; $x=2\Rightarrow y=0$x=2 y=0; $x=4\Rightarrow y=2$x=4 y=2. Points: $(0,-2),(2,0),(4,2)$(0,-2),(2,0),(4,2).

Intersection (algebraic check): adding the equations, $2x=8\Rightarrow x=4$2x=8 x=4, then $y=4-2=2$y=4-2=2. The lines cross at $(4,2)$(4,2), which is the simultaneous solution.

For $y=x$y=x: points $(0,0),(2,2),(4,4)$(0,0),(2,2),(4,4) — a line through the origin at $45^\circ$45^.

For $x+y=8$x+y=8 (i.e. $y=8-x$y=8-x): points $(0,8),(4,4),(8,0)$(0,8),(4,4),(8,0).

Both tables contain the point $(4,4)$(4,4), so the two lines cross there.

Check: in $y=x$y=x, $4=4$4=4 true; in $x+y=8$x+y=8, $4+4=8$4+4=8 true. Hence the solution is $x=4,\ y=4$x=4, y=4.

Write both in slope–intercept form $y=mx+c$y=mx+c.

$3x-2y=5\Rightarrow 2y=3x-5\Rightarrow y=\dfrac{3}{2}x-\dfrac52.$3x-2y=5 2y=3x-5 y=3/2x-52.

$6x-4y=7\Rightarrow 4y=6x-7\Rightarrow y=\dfrac{3}{2}x-\dfrac74.$6x-4y=7 4y=6x-7 y=3/2x-74.

Both have the same gradient $m=\dfrac32$m=32 but different $y$y-intercepts $\left(-\dfrac52\neq-\dfrac74\right)$(-52≠-74).

Equal gradient with unequal intercepts means the lines are parallel and distinct, so they never meet and the system has no solution (inconsistent).

Table for $x+y=4$x+y=4 ($y=4-x$y=4-x): $(0,4),(2,2),(4,0)$(0,4),(2,2),(4,0).

Table for $x-y=2$x-y=2 ($y=x-2$y=x-2): $(0,-2),(2,0),(3,1)$(0,-2),(2,0),(3,1).

Plotting both lines on the same axes:

The two lines cross at $(3,1)$(3,1).

Check: $3+1=4$3+1=4 and $3-1=2$3-1=2; both equations are satisfied.

$\therefore x=3,\ y=1.$x=3, y=1.

Let the numbers be $x$x and $y$y with $x>y$x>y.

Equations: $x+y=10$x+y=10 and $x-y=4$x-y=4.

Table for $x+y=10$x+y=10 ($y=10-x$y=10-x): $(2,8),(5,5),(10,0)$(2,8),(5,5),(10,0).

Table for $x-y=4$x-y=4 ($y=x-4$y=x-4): $(4,0),(5,1),(7,3)$(4,0),(5,1),(7,3).

Both tables lead to the point $(7,3)$(7,3); algebraically, adding the equations gives $2x=14\Rightarrow x=7$2x=14 x=7, then $y=10-7=3$y=10-7=3.

Check: $7+3=10$7+3=10 and $7-3=4$7-3=4; both true. So the numbers are $\boxed{7\text{ and }3}$7 and 3.

Where $x+y=10$x+y=10 meets the $x$x-axis: put $y=0\Rightarrow x=10$y=0 x=10, i.e. the point $(10,0)$(10,0).

(i) For $x+y=30$x+y=30, put $y=0\Rightarrow x=30$y=0 x=30: meets $x$x-axis at $(30,0)$(30,0). For $2x+y=40$2x+y=40, put $y=0\Rightarrow x=20$y=0 x=20: meets $x$x-axis at $(20,0)$(20,0).

(ii) Subtract the first equation from the second: $(2x+y)-(x+y)=40-30\Rightarrow x=10$(2x+y)-(x+y)=40-30 x=10. Then $10+y=30\Rightarrow y=20$10+y=30 y=20. The lines intersect at $(10,20)$(10,20).

(iii) So one pen costs $\rupee 10$10 and one notebook costs $\rupee 20$20.

Check: $10+20=30$10+20=30 and $2(10)+20=40$2(10)+20=40; both purchases match.