Q1Draw the graphs of the 5x-2y=10 and 3x+4y=12. shade the region bounded by these lines and x-axis. Also find the area of the shaded region.
Q2 Solve:
3/x+y+11/x-y=10
4/x+y+5/x-y=13
Q3 Solve:
8root x-15root y= -root xy
10/root x+8root y=4
where x and y are not equal to 0
Q4Show thatx=3 and y=2 is not a solution of the system of simultaneous linear equations 3x+4y=17 and 4x-3y=12
Answers
Answer:
Answer: 15/13 sq. units
Step-by-step explanation:
We have drawn two lines 5x-2y=10, 3x+4y=12 on the graph.
We got two three points which forms triangle A, B, C.
A is at y= 0 on line 5x-2y=10,
= > 5x=10
=> x = 2
=> A(2,0)
Now, B is on line 3x+4y=12 , at y=0 ,
=> x = 4
=>B (4,0)
Then, C is at point of intersection of both lines 3x+4y=12 and 5x-2y=10.
That is, Using Elimination method,
2*(5x-2y)=2*10
=>10x-4y=20
=>(3x+4y)= 12
+
= >13x=32
=>x=32/13
Substituting in any line out of two, we get y = 15/13 .
C(32/13,15/13)
Area of triangle =
\begin{lgathered}\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]\\ \\=\frac{1}{2}[2(0-15/13)+4(15/13-0)+32/13(0-0)]\\ \\=15/13\end{lgathered}
2
1
[x
1
(y
2
−y
3
)+x
2
(y
3
−y
1
)+x
3
(y
1
−y
2
)]
=
2
1
[2(0−15/13)+4(15/13−0)+32/13(0−0)]
=15/13
Hence, Area of triangle is 15/13 sq. units.
q2
Let 1/x+y be p
1/x-y be q
so the eqn is
3p+2q=3
2p+3q=11/3
Eliminate the equation
2[3p+2q=3] ------------1
3[2p+3q=11/3]----------2
subtract the equation
6p+4q=6
6p+9q=11
_ _ _
___________
-5q=-5
q=-5/-5
q=1
put the value of q in eqn 1
6p+4q=6
6p+4(1)=6
6p+4=6
6p=6-4
6p=2
p=2/6
p=1/3
1/x+y=p=1/3
1/x+y=1/3
3=x+y--------------3rd eqn
1/x-y=q=1
1/x-y=1
x-y=1---------4th eqn
x=1+y
put the value of x in eqn 3rd
3=x+y
3=1+y+y
3=1+2y
3-1=2y
2=2y
2/2=y
y=1
put the value of y in eqn 4th
x=1+y
x=1+1
x=2