Pair of Linear Equations in Two Variables
Mathematics Class Tenth
NCERT Exercise 3.3
Algebraic Methods of Solving a Pair of Linear Equations
Substitution Method
Let, a pair of linear equations in two variables are
`x+y=11` -------(i)
And `3x-2y = 3` -------(ii)
Steps to solve these linear equations by substitution method:
Step: 1. Find the value of one variable, say `y` in terms of the other variable, i.e., `x` from either equation, whichever is convenient.Example: From equation (i)
`x+y = 11`
`=> x = 11-y` ----------(iii)
Step: 2. Substitute this value of `x` in the other equation, and reduce it to an equation in one variable, i.e. in terms of `y`, and solve it and find the value of `y`.
Example: After substituting the value of `x` from equation (iii) in equation, (ii), we get
`3(11-y) - 2y = 3`
`=>33 -3y - 2y = 3`
`=> 33-5y = 3`
`=>-5y = 3 - 33`
`=> -5y = -30`
`:. y = (-30)/(-5) = 6`
Step: 3. Now, substitute the value of `y` in the equation (i) and calculate the value of `x`
Example:
After substituting the value of `y` in equation, (i) we get
`x+6=11`
`=> x = 11-6 = 5`
Thus, `x = 5` and `y = 6` Answer.
NCERT Exercise 3.3
Question: 1. Solve the following pair of linear equations by the substitution method.
(i) `x+y = 14`; `x-y = 4`
Solution:
Given, `x+y = 14` ------------(i)
`x-y = 4` -----------(ii)
Now, from equation (i) `x+y = 14`
`x= 14 - y` ---------(iii)
Now after substituting the value of `x` from equation (iii) in equation (ii), we get
`(14-y)-y = 4`
`=> 14-y - y = 4`
`=> 14-2y = 4`
`=> -2y = 4-14 = -10`
`=> y = (-10)/(-2) = 5`
Now, by substituting the value of `y` in equation (i), we get,
`x+5 = 14`
`=> x = 14-5 = 9`
Thus, `x = 9` and `y = 5` Answer
(ii) `s-t = 3`; `s/3+t/2 = 6`
Solution:
Given, `s-t = 3` --------(i)
`s/3+t/2 = 6` ---------(ii)
Now, `s-t = 3`
`=> s = 3+t` ------(iii)
And, `s/3 + t/2 = 6`
`=> (2s+3t)/6 = 6`
`=> 2s+3t = 6xx6`
`=> 2s+3t = 36`
After substituting the value of `s` from equation (iii), we get
`=> 2(3+t) + 3t = 36`
`=> 6+2t+3t = 36`
`=> 5t = 36-6 = 30`
`:. t = 30/5 = 6`
Now, after substituting the value of `t = 6` in equation (i), we get
`s - 6 = 3`
`=> s = 3+6 = 9`
Thus, `s = 9` and `t = 6` Answer
(iii) `3x-y = 3`; `9x-3y = 9`
Solution:
Given, `3x-y = 3` -------------(i)
`9x-3y = 9` -----------(ii)
Now, `3x-y = 3`
`=> 3x-3 = y`
`=> y = 3x-3` ----------(iii)
After substituting the value `y` from equation (iii) in equation (ii), we get
`9x - 3 (3x-3) = 9`
`=>\cancel( 9x) -\cancel( 9x)+9 =9`
`=>9 = 9` which is true
Here, from given pair of linear equations
`a_1 = 3, b_1 = -1, c_1 = -3`
And, `a_2 = 9, b_2 = -3, c_2 = -9`
Thus, `a_1/a_2 = 3/9 = 1/3`
`b_1/b_2 = (-1)/(-3) = 1/3`
`c_1/c_2 = (-3)/(-9) = 1/3`
Here, since, `a_1/a_2 = b_1/b_2 = c_1/c_2`
Thus, given pair of equation has infinitely many solutions.
From equation (i) `3x-y = 3`
If `x = 0`
`:. y = -3`
And, if `x = 1`
`=> 3xx1 - y = 3`
`=> - y = 3/3 = 1`
`=> y = -1`
Thus, one possible solution is `x = 0, y = -3` and `x = 1, y = -1` are two possible solutions.
(iv) `0.2x + 0.3y = 1.3`;
`0.4x + 0.5y = 2.3`
Solution:
Given, `0.2x + 0.3y = 1.3` ----------(i)
`0.4x + 0.5y = 2.3` ----------(ii)
Thus, now, equation (i)
`0.2x + 0.3y = 1.3`
`=> 0.2x = 1.3 - 0.3y `
`=> x = (1.3-0.3y)/(0.2)` ----------(iii)
Now, by substituting the value of `x` in equation (ii), we get
`0.4xx(1.3-0.3y)/(0.2) + 0.5y = 2.3`
`=> (0.52 - 0.12y)/(0.2)+0.5y = 2.3`
`=> ((0.52-0.12y)+0.1y)/(0.2) = 2.3`
`=> 0.52-0.12y +0.1y = 2.3xx0.2`
`=> 0.52 - 0.02y = 0.46`
`=> -0.02 y = 0.46 - 0.52`
`=> - 0.02 y = - 0.06`
`=> y = (0.06)/(0.02) = 3`
Now, after substituting the value of `y` in equation (i), we get
`0.2x + 0.3 xx 3 = 1.3`
`=> 0.2x + 0.9 =1.3`
`=> 0.2x = 1.3 - 0.9`
`=> x = (0.4)/(0.2)`
`=> x = 2`
Thus, `x = 2` and `y = 3` Answer
(v) `sqrt2\x+sqrt3\y = 0`;
`sqrt3\x-sqrt8\y=0`
Solution:
Given,
`sqrt2\x+sqrt3\y = 0` --------- (i)
`sqrt3\x-sqrt8\y=0` ----------- (ii)
Now, `sqrt2\x+sqrt3\y = 0`
`=> sqrt2\x = -sqrt3\y = 0`
`=x = -(sqrt3\y)/sqrt2` ---------(iii)
After substituting the value of `x` in equation (ii), we get
`sqrt3 xx (-sqrt3\y)/sqrt2 - sqrt8\y = 0`
`=>(-3y)/sqrt2 - sqrt8\y =0`
`=>(-3y - sqrt2xxsqrt8\y)/sqrt2 =0`
`=> - 3y - sqrt16\y = 0`
`=> y (-3-4) = 0`
`=> y = 0/(-1) = 0`
Now, by substituting the value of `y` in equation (i)
`sqrt2\x - sqrt3 xx 0 = 0`
`=>sqrt2\x - 0=0`
`=> x = 0`
Thus, `x = 0` and `y = 0` Answer
(vi) `(3x)/2 - (5y)/3 = -2`;
`x/3 + y /2 = 13/6`
Solution:
Given, `(3x)/2 - (5y)/3 = -2` -------(i)
`x/3 + y /2 = 13/6` --------------(ii)
Now, from equation (i)
`(3x)/2 - (5y)/3 = -2`
`=> (9x-10y)/6 = -2`
`=> 9x-10y = -12`
`=> 9x = (-12 + 10y)`
`=> x = (-12+10y)/9` -----------(iii)
After substituting the value of `x` in equation (ii), we get
`((-12+10y)//9)/3 +y/2 = 13/6`
`=> (-12+10y)/27 + y/2 = 13/6`
`=>(2xx(-12+10y) + 27 y)/54 = 13/6`
`=> -24 + 20y + 27 y = (13xx\cancel(54)9)/\cancel(6)`
`=> -24 + 47 y = 117`
`=> 47y = 117+24= 141`
`=> y = 141/47 = 3`
Now, by substituting the value of `y` in equation (ii), we get
`x/3 + 3/2 = 13/6`
`=>(2x+9)/6=13/6`
`=>2x+9 = (13xx6)/6`
`=> 2x+9 = 13`
`=> 2x = 13-9 =4`
` :. x = 4/2 = 2`
Thus, `x = 2` and `y = 3` Answer
Question: 2. Solve `2x+3y =11` and `2x-4y= -24` and hence find the value of `m` for which `y = mx +3`.
Solution:
Given, `2x+3y = 11` -----------(i)
And, `2x-4y = -24` ------------(ii)
Now, from equation (i)
`2x+3y = 11`
`=> 2x = 11-3y`
`=> x = (11-3y)/2` -------------(iii)
Now, by substituting the value of `x` from equation (iii) in equation (ii), we get
`2((11-3y)/2)-4y = -24`
`=> 11-3y - 4y = -24`
`=> 11-7y = -24`
`=> -7y = -24 -11`
`=>-7 y = -35`
`=>y = (-35)/(-7) = 5`
Now, by substituting the value `y` in equation (i), we get
`2x + 3xx5 = 11`
`=>2x +15 = 11`
`=>2x = 11-15 = -4`
`=> x = (-4)/2 = -2`
Thus, `x = -2` and `y = 5`
Now, as given, `y = mx +3` -----------(iv)
After substituting the values of `x` and `y` in equation (iv), we get
`5 = m(-2) +3`
`=> -2m = 5-3 = 2`
`=> m = 2(-2) = -1`
Thus, `m = -1` Answer
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