1: A) Fill in the blanks.
1) The way of matching the vertices is called
7m
2) If
+5=12 then m = ?
8
3) If the mean of 8, 6, 9, 5, P, 3 and 7 is 6 then P=
4) 35m + m =
D) Solve any five.
1) In a paper of 40 marks Jay got 36 marks and Viru got 109
marks then find the marks obtained by Viru?
2) Find the value of x
3x 2x
Answers
Answer:
!!!
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Answer:
Mathematics NCERT Grade 7, Chapter 4: Simple equations- The beginning of the chapter is given in a very interesting way. It starts by citing a mind-reading game. A little recap of pre-known knowledge about the chapter is given in the chapter. Here the discussion is basically on how a statement is converted into an expression followed by an equation using variables.
The first portion of the chapter discusses the following topic:
What Equation Is?
An equation is a condition on a variable such that two expressions in the variable should have equal value.
Once the student has learned about how an equation is formed from a variable, the next topic is Solving an Equation.
The value of the variable for which the equation is satisfied is called the solution of the equation.
There are two methods for find the solution of the equation, once is by adding or subtracting number on both sides and the other one is transposing. Transposing a number is the same as adding or subtracting the number from both sides.
An equation remains the same if the Left hand side(LHS) and the right hand side(RHS) are interchanged.
Equation --------> Solution (normal path)
Solution ---------> Equation (reverse path)
This shows that we can not only can solve an equation, but also can make equations by following the reverse path. Also, given an equation, we can get one solution but with the given solution we can make many equations.
Applications of Simple Equations To Practical Situations is the last topic of the chapter. In this section, students will learn how to construct simple equations corresponding to practical situations and thus finding the solution of the equations using suitable methods like transposing.
Exercise 4.4 is the last exercise given followed by the summary of the chapter- Simple Equations.
(i) x + 3 = 0
L.H.S. = x + 3
By putting x = 3,
L.H.S. = 3 + 3 = 6 ≠ R.H.S.
∴ No, the equation is not satisfied.
(ii) x + 3 = 0
L.H.S. = x + 3
By putting x = 0,
L.H.S. = 0 + 3 = 3 ≠ R.H.S.
∴ No, the equation is not satisfied.
(iii) x + 3 = 0
L.H.S. = x + 3
By putting x = −3,
L.H.S. = − 3 + 3 = 0 = R.H.S.
∴ Yes, the equation is satisfied.
(iv) x − 7 = 1
L.H.S. = x − 7
By putting x = 7,
L.H.S. = 7 − 7 = 0 ≠ R.H.S.
∴ No, the equation is not satisfied.
(v) x − 7 = 1
L.H.S. = x − 7
By putting x = 8,
L.H.S. = 8 − 7 = 1 = R.H.S.
∴ Yes, the equation is satisfied.
(vi) 5x = 25
L.H.S. = 5x
By putting x = 0,
L.H.S. = 5 × 0 = 0 ≠ R.H.S.
∴ No, the equation is not satisfied.
(vii) 5x = 25
L.H.S. = 5x
By putting x = 5,
L.H.S. = 5 × 5 = 25 = R.H.S.
∴ Yes, the equation is satisfied.
(viii) 5x = 25
L.H.S. = 5x
By putting x = −5,
L.H.S. = 5 × (−5) = −25 ≠ R.H.S.
∴ No, the equation is not satisfied.
(ix) = 2
L.H.S. =
By putting m = −6,
L. H. S. = ≠ R.H.S.
∴No, the equation is not satisfied.
(x) = 2
L.H.S. =
By putting m = 0,
L.H.S. = ≠ R.H.S.
∴No, the equation is not satisfied.
(xi) = 2
L.H.S. =
By putting m = 6,
L.H.S. = = R.H.S.
∴ Yes, the equation is satisfied.
Step-by-step explanation: