Page 63 - SM inner class 8.cdr

5(x - 2) + (x - 3) = 2(2x + 1) - 9 are equal to 3x + 4
L.H.S. =
each other. 5
Verification: Putting x = 27, we get;
´
+
L.H.S. = 5(x - 2) + (x - 3) = (3 27 4 )
Putting the value of x = 3 we get; 5
4
= 5(3 - 2) + (3 - 3) = 81 +
5
= 5 + 0 = 5
85
R.H.S. = 2(2x + 1) - 9 = = 17
5
Putting the value of x = 3, we get
Verification:
= 2(2 ´ 3 + 1) - 9 2x - 3
R.H.S. =
= 2(6 + 1) - 9 3
= 2 ´ (7) - 9 Putting x = 27, we get;
= 14 - 9 = 5 (2 27 ) 3
´
-
=
Since, L.H.S. = R.H.S. hence verified. 3
3
Cross Multiplication Method = 54 -
3
The process of multiplying the numerator on 51
= = 17
the left hand side with the denominator on the 3
right hand side and multiplying the
Since, L.H.S. = R.H.S. hence verified.
denominator on left hand side with the
Example 5: Solve 0.8 - 0.28x = 1.16 - 0.6x
numerator on right hand side is called cross
multiplication. Solution:

And then equating both the products we get 0.8 - 0.28x = 1.16 - 0.6x
the linear equation. Þ 0.6x - 0.28x = 1.16 - 0.8
ax + b p Þ 0.32x = 0.36
(
If = then, q ax( + b) = p cx + d).
.
cx + d q 036
Þ x =
.
On solving it we get the value of variable for 032
36
which L.H.S. = R.H.S. Then, it is an equation Þ x =
of the form. 32
9
Þ x =
Cross multiplication while solving linear 8
equations: 9
Therefore, is the required solution.
3x + 4 2x - 3 8
Example 4: =
5 3 Verification:
3x + 4 2x -3
Solution: = L.H.S. = 0.8 - 0.28x
5 3 9
Putting x = , we get;
On cross multiplication, we get; 8
3 (3x + 4) = 5(2x - 3) 9
= 0.8 - 0.28 ×
Þ 9x + 12 = 10x - 15 8
Þ 9x - 10x = -15 - 12 = 8 - 28 ´ 9
Þ -x = -27 10 100 8
8 63
Þ x = 27 = -
10 200
Verification:

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Mathematics In Focus - 8