Question

How do you solve \[80=3y+2y+4+1\] ?

Answer 1

Hint: For the given problem we have to solve the equation \[80=3y+2y+4+1\] and we have to find the value of y. For this first of all we have to add all constant terms in RHS as well as we have to add all variable terms in RHS. After that we have to send all constant terms a side and variables a side. Then we have to solve the problem for the value of y.

Complete step-by-step answer:
For the given problem, we have to solve \[80=3y+2y+4+1\] and we have to find the value of y.
Now we have to consider the equation as equation (1)
Let us consider the above equation as equation (1).
\[80=3y+2y+4+1............\left( 1 \right)\]
Now we have to add the terms 4, 1.
After adding the terms we have
\[80=3y+2y+5\]
Let us consider the above equation as equation (2).
\[80=3y+2y+5.............\left( 2 \right)\]
Now we have to add terms 3y, 2y.
After adding the terms, we get
\[80=5y+5\]
Let us consider the above equation as equation (3).
\[80=5y+5...........\left( 3 \right)\]
Transferring 5 from RHS (Right hand side) to LHS (Left hand side).
\[80-5=5y\]
By subtracting 5 from 80 we get
\[75=5y\]
Let us consider the above equation as equation (4).
\[75=5y..........\left( 4 \right)\]
Now Transferring 5 from RHS (Right hand side) to LHS (Left hand side).
\[\dfrac{75}{5}=y\]
Dividing 75 by 5, we get
\[15=y\]
Let us consider the above equation as equation (5)
\[15=y......\left( 5 \right)\]
Rewriting the equation by changing the sides.
\[y=15\]
Let us consider the above equation as equation (6).
\[y=15..........\left( 6 \right)\]
Therefore, the solution of the equation \[80=3y+2y+4+1\] is\[y=15\].

Note: We should have a knowledge on solving algebraic expressions to solve this problem. Students should practise some difficult problems to be perfect in this concept. Sometimes the examiner may give a quadratic equation in y and asks to solve. So, students should have keen knowledge on solving algebraic expressions.