**4 5 P 1 34**. How do you solve p−1 = 4p +17 ? P− 1 = 4p+17 −1−17 = 4p− p −18 = 3p p = − 318. Solve the following equation 34−5(p−1)=4 easy solution verified by toppr 34−5(p−1)=4⇒−5(p−1)=4−34⇒−5(p−1)=−30 ⇒p−1= −5−30⇒p−1=6⇒p=6+1 ⇒p=7 was this. 11, 2021 by teachoo transcript ex 4.3, 3 solve the following equations: The questions and answers of 4 5 ( p=1) =34? We use the concepts of simple linear equations and fundamental operations like addition , subtraction , multiplication and division to. (d) 4 + 5 ( p − 1) = 34 ⇒ 4 + 5 p − 5 = 34 ⇒ 5 p − 1 = 34 ⇒ 5 p = 34 + 1 ⇒ 5 p = 35 ⇒ p = 35 5 ⇒ p = 7 (e) 0 = 16 + 4 ( m − 6) ⇒ 0 = 16 + 4 m − 24 ⇒ 0 = 4 m − 8 ⇒ 8 = 4 m ⇒ m.

Are solved by group of students and teacher of. 4+5(p−1)=34 easy solution verified by toppr we have, 4+5(p−1)=34 4+5p−5=34 5p−1=34 5p=34+1 5p=35 p=7 hence, this is the answer. How do you solve p−1 = 4p +17 ? 11, 2021 by teachoo transcript ex 4.3, 3 solve the following equations: We use the concepts of simple linear equations and fundamental operations like addition , subtraction , multiplication and division to. Is done on edurev study group by class 7 students. Two angles are supplementary angle a is twice as large;

## /* divide both sides by 5 , we get */ $\implies.

4+5(p−1)=34 easy solution verified by toppr we have, 4+5(p−1)=34 4+5p−5=34 5p−1=34 5p=34+1 5p=35 p=7 hence, this is the answer. How do you solve p−1 = 4p +17 ? Are solved by group of students and teacher of. We have to find the value of p. P = 6 + 1 = 7 (transposing −1 to r.h.s.) concept: What is the formulae of. This discussion on 4 5 ( p=1) =34? Two angles are supplementary angle a is twice as large;

### /* Divide Both Sides By 5 , We Get */ $\Implies.

We use the concepts of simple linear equations and fundamental operations like addition , subtraction , multiplication and division to. We have to find the value of p. What is the formulae of. 4+5(p−1)=34 easy solution verified by toppr we have, 4+5(p−1)=34 4+5p−5=34 5p−1=34 5p=34+1 5p=35 p=7 hence, this is the answer. Are solved by group of students and teacher of. 11, 2021 by teachoo transcript ex 4.3, 3 solve the following equations:

## /* Divide Both Sides By 5 , We Get */ $\Implies.

Solve the following equation 34−5(p−1)=4 easy solution verified by toppr 34−5(p−1)=4⇒−5(p−1)=4−34⇒−5(p−1)=−30 ⇒p−1= −5−30⇒p−1=6⇒p=6+1 ⇒p=7 was this. Two angles are supplementary angle a is twice as large; We have to find the value of p. P− 1 = 4p+17 −1−17 = 4p− p −18 = 3p p = − 318. This discussion on 4 5 ( p=1) =34? (d) 4 + 5 ( p − 1) = 34 ⇒ 4 + 5 p − 5 = 34 ⇒ 5 p − 1 = 34 ⇒ 5 p = 34 + 1 ⇒ 5 p = 35 ⇒ p = 35 5 ⇒ p = 7 (e) 0 = 16 + 4 ( m − 6) ⇒ 0 = 16 + 4 m − 24 ⇒ 0 = 4 m − 8 ⇒ 8 = 4 m ⇒ m.

## Conclusion of **4 5 P 1 34**.

. Solve the following equation 34−5(p−1)=4 easy solution verified by toppr 34−5(p−1)=4⇒−5(p−1)=4−34⇒−5(p−1)=−30 ⇒p−1= −5−30⇒p−1=6⇒p=6+1 ⇒p=7 was this. Is done on edurev study group by class 7 students.

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