Question

How do you isolate r in $4\left( x-3 \right)-r+2x=5\left( 3x-7 \right)-9x$ ?

Answer 1

Hint: We have to isolate r in the equation given in the question. We can see -r is at LHS in the equation of we can send – r to RHS and send all the terms in RHS to LHS. Then only we can isolate r in the equation.

Complete step by step answer:
The given equation in the question is $4\left( x-3 \right)-r+2x=5\left( 3x-7 \right)-9x$
We will isolate r by taking all the x and all the constants to the LHS and send r to RHS
When we send some value from LHS to RHS or RHS to LHS we just change the sign from plus to minus, minus to plus, multiplication to division and division to multiplication of the term that means if there is a term – m in LHS then it will be m in RHS.
If we observe we can see – r present in LHS so we can send it to RHS so it will become r
So we can write $4\left( x-3 \right)+2x=5\left( 3x-7 \right)+r-9x$
Now we can see the term -9x and $5\left( 3x-7 \right)$ in RHS we can send both the terms to LHS , so -9x become 9x and $5\left( 3x-7 \right)$ become $-5\left( 3x-7 \right)$
$-5\left( 3x-7 \right)+4\left( x-3 \right)+2x+9x=r$
Further solving the equation we get
$23=r$

Note: We know when we send terms from RHS to LHS or LHS to RHS the sign of the term changes but there is an exception , where a number is multiplied with 0 in one side we can not send 0 to other side by diving 0 .We can not divide something with 0.