# NCERT Class 10 Mathematics Linear Equation Formative Assessment CBSE Board Sample Problems Part 1 (For CBSE, ICSE, IAS, NET, NRA 2022)

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## Linear Equation Formative Assessment

Question 1. Which of these equation have i) Unique solution ii) Infinite solutions iii) no solutions

a)

b)

c)

d)

e)

f)

g)

h)

Solution

Condition | Algebraic interpretation |

One unique solution only | |

Infinite solution | |

No solution |

One unique solution: (a) , (d) , (e) , (f) , (d)

Infinite solution: (b)

No solution: (e) , (h)

Question 2. Using substitution method solve the below equation

a)

b)

Solution

1 | Method of elimination by substitution | 1) Suppose the equation are 2) Find the value of variable of either x or y in other variable term in first equation 3) Substitute the value of that variable in second equation 4) Now this is a linear equation in one variable. Find the value of the variable 5) Substitute this value in first equation and get the second variable |

a) From 1^{st} equation

Substituting this in second equation

Putting this 1^{st} equation

b) (1,0)

Question 3. Using elimination method, solve the following

a)

b)

Solution

1 | Method of elimination by equating the coefficients | 1) Suppose the equation are 2) Find the LCM of and . Let it k. 3) Multiple the first equation by the value 4) Multiple the first equation by the value 4) Subtract the equation obtained. This way one variable will be eliminated and we can solve to get the value of variable y 5) Substitute this value in first equation and get the second variable |

a) … (1)

… (2)

Multiplying equation (1) by 7

… (3)

Subtracting equation 2 from equation 7

We get

Substituting this in (1) , we get x = 15

b) (8,6)

Question 4. Solve the below linear equation using cross-multiplication method

a)

b)

Solution

3 | Cross Multiplication method | 1) Suppose the equation are 2) This can be written as 3) This can be written as 4) Value of x and y can be find using the first and last expression second and last expression |

a) The equation can be written as

By cross multiplication we have

Solving

b)