Linear Equations
NCERT Exercise 3.3
Part 1
Solving by substitution method:
- One of the equations is picked and in this, one variable is expressed in terms of another variable.
- The value of one variable (in terms of another variable) is substituted in the second equation to calculate the value of the remaining variable.
Question 1: Solve the following pair of linear equations by the substitution method.
(a) x + y = 14 and x – y = 4
Solution: Let us use the first equation to express one variable in terms of another variable;
`x + y = 14`
Or, `x = 14 – y`
Substituting the value of x in second equation we get:
`x – y = 4`
Or, `14 – y – y = 4`
Or, `14 – 2y = 4`
Or, `2y = 14 – 4 = 10`
Or, `y = 5`
Substituting the value of y in first equation, we get;
`x = 14 – y`
Or, `x = 14 – 5 = 9`
Hence, `x = 9` and `y = 5`
(b) `s-t=3` and `(s)/(3)+(t)/(2)=6`
Solution: Let us use the first equation to express one variable in terms of another variable;
`s – t = 3`
Or, `s = t + 3`
Substituting the value of s in second equation, we get;
`(s)/(3)+(t)/(2)=6`
Or, `(t+3)/(3)+(t)/(2)=6`
Or, `5t+6=36`
Or, `5t=36-6=30`
Or, `t=30÷5=6`
Substituting the value of t in first equation, we get;
`s = t + 3`
Or, `s = 6 + 3 = 9`
Hence, `s = 9` and `t = 6`
(c) `3x – y = 3` and `9x – 3y = 9`
Solution: Let us use first equation to express one variable in terms of another variable.
`3x – y = 3`
Or, `y = 3x – 3`
The second equation is same as the first equation;
`9x – 3y = 9`
Dividing the equation by 3, we get;
`3x – y = 3`
Hence, the given pair of linear equations has infinitely many solutions.
(d) `0.2x + 0.3y = 1.3` and `0.4x + 0.5y = 2.3`
Solution: Let us use the first equation to express one variable in terms of another variable;
`0.2x + 0.3y = 1.3`
Or, `2x + 3y = 13`
Or, `2x = 13 – 3y`
Or, `x = (13 – 3y)/(2)`
Substituting the value of x in second equation, we get;
`0.4x+0.5y=2.3`
Or, `4x+5y=23`
Or, `4((13-3y)/(2))+5y=23`
Or, `26-6y+5y=23`
Or, `26-y=23`
Or, `y=26-23=3`
Substituting the value of y in first equation, we get;
`x=(13-3y)/(2)`
`=(13-3xx3)/(2)=(13-9)/(2)`
`=4/2=2`
Hence, `x = 2` and `y = 3`
(e) `sqrt2x + sqrt3y = 0` and `sqrt3x - sqrt8y = 0`
Solution: Let us use first equation to express one variable in terms of another variable;
`sqrt2x+sqrt3y=0`
Or, `sqrt2x=-sqrt3y`
Or, `x=-(sqrt3)/(sqrt2)y`
Substituting the value of x in second equation, we get;
`sqrt3x-sqrt8y=0`
Or, `sqrt3x=sqrt8y`
Or, `-(sqrt3)/(sqrt2)y\xx\sqrt3=sqrt9y`
Or, `-(3)/(sqrt2)y=sqrt8y`
Or, `-3y=sqrt(16)y`
Or, `-3y=4y`
Or, `y=-(3)/(4)y`
Or, `y=-(4)/(3)y`
This can be possible only when the value of y is zero. Substituting the value of y in first equation, we get;
`x=-(sqrt3)/(sqrt2)y` Or, `x=0`
Hence, x = 0 and y = 0
(f) `(3x)/(2)-(5y)/(3)=-2` and `x/3+y/2=(13)/(6)`
Solution: Let us use first equation to express one variable in terms of another variable;
`(3x)/(2)-(5y)/(3)=-2`
Or, `(9x-10y)/(6)=-2`
Or, `9x-10y=-12`
Or, `9x+12=10y`
Or, `y=(9x+12)/(10)`
Substituting the value of y in second equation, we get;
`x/3+y/2=(13)/(6)`
Or, `x/3+(9x+12)/(20)=(13)/(6)`
Or, `(20x+27x+36)/(60)=(13)/(6)`
Or, `47x+36=130`
Or, `47x=130-36=94`
Or, `x=94÷47=2`
Substituting the value of x in first equation, we get;
`y=(9x+12)/(10)`
Or, `y=(18+12)/(10)=(30)/(10)=3`
Hence, `x = 2` and `y = 3`
Question 2: Solve `2x + 3y = 11` and `2x – 4y = - 24` and hence find the value of ‘m’ for which `y = mx + 3`.
Solution: Let us use first equation to express one variable in terms of another variable;
`2x + 3y = 11`
Or, `2x = 11 – 3y`
Substituting the value of x in second equation, we get;
`2x – 4y = - 24`
Or, `11 – 3y – 4y = - 24`
Or, `11 – 7y = - 24`
Or, `7y = 11 + 24 = 35`
Or, `y = 5`
Substituting the value of y in first equation, we get;
`2x = 11 – 3y`
Or, `2x = 11 – 3 xx 5`
Or, `2x = 11 – 15 = - 4`
Or, `x = - 2`
Hence, `x = - 2` and `y = 5`
Now; we have to find the value of m in following equation;
`y = mx + 3`
Or, `5 = m( - 2) + 3`
Or, `- 2m + 3 = 5`
Or, `- 2m = 5 – 3 = 2`
Or, `m = - 1`