
How do you solve $ 7 = \dfrac{3}{5}t - 2 $ ?
Answer
548.1k+ views
Hint: In order to determine the value of variable $ t $ in the above equation. Use the rules of transposing terms to transpose terms having $ t $ on the right-hand side and constant value terms on the left-Hand side of the equation. Combine and solve like terms and multiply the equation with a reciprocal of coefficient of $ t $ to get your desired solution.
Complete step-by-step answer:
We are given a linear equation in one variable $ 7 = \dfrac{3}{5}t - 2 $ .and we have to solve this equation for variable ( $ t $ ).
$ \Rightarrow 7 = \dfrac{3}{5}t - 2 $
Now combining like terms on both of the sides. Terms having $ t $ will on the right-Hand side of the equation and constant terms on the left-hand side.
Let’s recall one basic property of transposing terms that on transposing any term from one side to another the sign of that term gets reversed .In our case, $ - 2 $ on the right-hand side will become $ 2 $ on the left-hand side .
After transposing terms our equation becomes
$
\Rightarrow 7 + 2 = \dfrac{3}{5}t \\
\Rightarrow 9 = \dfrac{3}{5}t \;
$
Multiplying both sides of the equation with the number $ \dfrac{5}{3} $ ,we get
$
\Rightarrow \dfrac{5}{3} \times 9 = \dfrac{5}{3} \times \dfrac{3}{5}t \\
\Rightarrow 15 = t \\
\Rightarrow t = 15 \;
$
Therefore, the solution to the equation $ 7 = \dfrac{3}{5}t - 2 $ is equal to $ t = 15 $ .
So, the correct answer is “ $ t = 15 $ ”.
Note: Linear Equation in one variable: A linear equation is a equation which can be represented in the form of $ ax + c $ where $ x $ is the unknown variable and a,c are the numbers known where $ a \ne 0 $ .If $ a = 0 $ then the equation will become constant value and will no more be a linear equation .
The degree of the variable in the linear equation is of the order 1.
Every Linear equation has 1 root.
Also,
1. One must be careful while calculating the answer as calculation error may come.
2.There is only one value of $ t $ which is the solution to the equation and if we put this $ t $ in the equation, the equation will be zero.
3.Like terms are the terms who have the same variable and power.
Complete step-by-step answer:
We are given a linear equation in one variable $ 7 = \dfrac{3}{5}t - 2 $ .and we have to solve this equation for variable ( $ t $ ).
$ \Rightarrow 7 = \dfrac{3}{5}t - 2 $
Now combining like terms on both of the sides. Terms having $ t $ will on the right-Hand side of the equation and constant terms on the left-hand side.
Let’s recall one basic property of transposing terms that on transposing any term from one side to another the sign of that term gets reversed .In our case, $ - 2 $ on the right-hand side will become $ 2 $ on the left-hand side .
After transposing terms our equation becomes
$
\Rightarrow 7 + 2 = \dfrac{3}{5}t \\
\Rightarrow 9 = \dfrac{3}{5}t \;
$
Multiplying both sides of the equation with the number $ \dfrac{5}{3} $ ,we get
$
\Rightarrow \dfrac{5}{3} \times 9 = \dfrac{5}{3} \times \dfrac{3}{5}t \\
\Rightarrow 15 = t \\
\Rightarrow t = 15 \;
$
Therefore, the solution to the equation $ 7 = \dfrac{3}{5}t - 2 $ is equal to $ t = 15 $ .
So, the correct answer is “ $ t = 15 $ ”.
Note: Linear Equation in one variable: A linear equation is a equation which can be represented in the form of $ ax + c $ where $ x $ is the unknown variable and a,c are the numbers known where $ a \ne 0 $ .If $ a = 0 $ then the equation will become constant value and will no more be a linear equation .
The degree of the variable in the linear equation is of the order 1.
Every Linear equation has 1 root.
Also,
1. One must be careful while calculating the answer as calculation error may come.
2.There is only one value of $ t $ which is the solution to the equation and if we put this $ t $ in the equation, the equation will be zero.
3.Like terms are the terms who have the same variable and power.
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