Math, asked by daar3054, 5 months ago

The statement form of the equation x/2 +6=7 is : ​

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Answered by jkbkumr1234543
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Answered by amreet28
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Answer:

Class 7 - Maths - Simple Equations

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Exercise 4.1

Question 1:

Complete the last column of the table:

Class_7_Simple_Equations_Forming_Equation

Answer:

(i) x + 3 = 0

LHS = x + 3

Put x = 3, we get

LHS = 3 + 3 = 6 ≠ RHS

So, the equation is not satisfied.

(ii) x + 3 = 0

LHS = x + 3

Put x = 0, we get

LHS = 0 + 3 = 3 ≠ RHS

So, the equation is not satisfied.

(iii) x + 3 = 0

LHS = x + 3

Put x = -3, we get

LHS = -3 + 3 = 0 = RHS

So, the equation is satisfied.

(iv) x - 7 = 1

LHS = x - 7

Put x = 7, we get

LHS = 7 - 7 = 0 ≠ RHS

So, the equation is not satisfied.

(v) x - 7 = 1

LHS = x - 7

Put x = 8, we get

LHS = 8 - 7 = 1 = RHS

So, the equation is satisfied.

(vi) 5x = 25

LHS = 5x

Put x = 0, we get

LHS = 5 * 0 = 0 ≠ RHS

So, the equation is not satisfied.

(vii) 5x = 25

LHS = 5x

Put x = 5, we get

LHS = 5 * 5 = 25 = RHS

So, the equation is satisfied.

(viii) 5x = 25

LHS = 5x

Put x = -5, we get

LHS = 5 * (-5) = -25 ≠ RHS

So, the equation is not satisfied.

(ix) m/3 = 2

LHS = m/3

Put m = -6, we get

LHS = -6/3 = -2 ≠ RHS

So, the equation is not satisfied.

(x) m/3 = 2

LHS = m/3

Put m = 0, we get

LHS = 0/3 = 0 ≠ RHS

So, the equation is not satisfied.

(xi) m/3 = 2

LHS = m/3

Put m = 6, we get

LHS = 6/3 = 2 = RHS

So, the equation is satisfied.

Question 2:

Check whether the value given in the brackets is a solution to the given equation or not:

(a) n + 5 = 19 (n = 1) (b) 7n + 5 = 19 (n = -2) (c) 7n + 5 = 10 (n = 2)

(d) 4p – 3 = 13 (p = 1) (e) 4p – 3 = 13 (p = -4) (f) 4p – 3 = 13 (p = 0)

Answer:

(a) n + 5 = 19 (n = 1)

LHS = n + 5

Put n = 1 in LHS, we get

LHS = 1 + 5 = 6 ≠ RHS

So, n = 1 is not the solution of given equation.

(b)7n + 5 = 19 (n = -2)

LHS = 7n + 5

Put n = -2 in LHS, we get

LHS = 7 * (-2) + 5 = -14 + 5 = -9 ≠ RHS

So, n = -2 is not the solution of given equation.

(a) 7n + 5 = 19 (n = 2)

LHS = 7n + 5

Put n = 2 in LHS, we get

LHS = 7 * 2 + 5 = 14 + 5 = 19 = RHS

So, n = 2 is the solution of given equation.

(d) 4p – 3 = 13 (p = 1)

LHS = 4p - 13

Put p = 1 in LHS, we get

LHS = 4 * 1 - 3 = 4 - 3 = 1 ≠ RHS

So, p = 1 is not the solution of given equation.

(e) 4p – 3 = 13 (p = -4)

LHS = 4p - 13

Put p = -4 in LHS, we get

LHS = 4 * (-4) - 3 = -16 - 3 = -19 ≠ RHS

So, p = -4 is not the solution of given equation.

(f) 4p – 3 = 13 (p = 0)

LHS = 4p - 13

Put p = 0 in LHS, we get

LHS = 4 * 0 - 3 = 0 - 3 = -3 ≠ RHS

So, p = 0 is not the solution of given equation.

Question 3:

Solve the following equations by trial and error method:

(i) 5p + 2 = 17 (ii) 3m – 14 = 4

Answer:

(i) 5p + 2 = 17

Put p = 1 in LHS

5 * 1 + 2 = 5 + 2 = 7 ≠ RHS

Put p = 2 in LHS

5 * 2 + 2 = 10 + 2 = 12 ≠ RHS

Put p = 3 in LHS

5 * 3 + 2 = 15 + 2 = 17 = RHS

Hence, p = 3 is the solution of the given method.

(ii) 3m – 14 = 4

Put m = 4 in LHS

3 * 4 – 14 = 12 – 14 = -2 ≠ RHS

Put m = 5 in LHS

3 * 5 – 14 = 15 – 14 = 1 ≠ RHS

Put m = 6 in LHS

3 * 6 – 14 = 18 – 14 = 4 = RHS

Hence, m = 6 is the solution of the given method.

Question 4:

Write equations for the following statements:

(i) The sum of numbers x and 4 is 9.

(ii) 2 subtracted from y is 8.

(iii) Ten times a is 70.

(iv) The number b divided by 5 gives 6.

(v) Three-fourth of t is 15.

(vi) Seven times m plus 7 gets you 77.

(vii) One-fourth of a number x minus 4 gives 4.

(viii) If you take away 6 from 6 times y, you get 60.

(ix) If you add 3 to one-third of z, you get 30.

Answer:

(i) The sum of numbers x and 4 is 9.

=> x + 4 = 9

(ii) 2 subtracted from y is 8.

=> y – 2 = 8

(iii) Ten times a is 70.

=> 10a = 70

(iv) The number b divided by 5 gives 6.

=> b/5 = 6

(v) Three-fourth of t is 15.

=> 3t/4 = 15

(vi) Seven times m plus 7 gets you 77.

=> 7m + 7 = 77

(vii) One-fourth of a number x minus 4 gives 4.

=> x/4 – 4 = 4

(viii) If you take away 6 from 6 times y, you get 60.

=> 6y – 6 = 60

(ix) If you add 3 to one-third of z, you get 30.

=> z/3 + 3 = 30

Question 5:

Write the following equations in statement form:

(i) p + 4 = 15 (ii) m – 7 = 3 (iii) 2m = 7 (iv) m/5 = 3

(v) 3m/5 = 6 (vi) 3p + 4 = 25 (vii) 4p – 2 = 18 (viii) p/2 + 2 = 8

Answer:

(i) p + 4 = 15

The sum of numbers p and 4 is 15.

(ii) m – 7 = 3

7 subtracted from m is 3.

(iii) 2m = 7

Two times m is 7.

(iv) m/5 = 3

The number m is divided by 5 gives 3.

(v) 3m/5 = 6

Three-fifth of the number m is 6.

(vi) 3p + 4 = 24

Three times p plus 4 gets 25.

(vii) 4p – 2 = 18

If you take away 2 from 4 times p, you get 18.

(viii) p/2 + 2 = 8

If you added 2 to half is p, you get 8.

Question 6:

Set up an equation in the following cases:

(i) Irfan says that he has 7 marbles more than five times the marbles Parmit has. Irfan has 37 marbles.

(Tale m to be the number of Parmit’s marbles.)

(ii) Laxmi’s father is 49 years old. He is 4 years older

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