How do you solve $13=\dfrac{w-14}{2}$ ?
Answer
629.1k+ views
Hint: We recall the linear equations and basic properties of linear equations. We solve the given linear equation by eliminating the constant terms at the right hand side so that only the variable term of $w$ remains at the right hand side. We multiply 2 both sides of the given equation and then add 14 both sides.
Complete step-by-step answer:
We know from algebra that the linear equation in one variable $x$ and constants $a\ne 0,b$is given by
\[ax=b\]
The term with which the variable is multiplied is called variable term and with whom not multiplied is called a constant term. We also know that if we add, subtract, multiply or divide the same number on both sides of the equation, equality holds. It is called balancing the equation. It means for some term $c$ we have,
\[\begin{align}
& ax+c=b+c \\
& ax-c=b-c \\
& ax\times c=b\times c \\
& \dfrac{ax}{c}=\dfrac{b}{c} \\
\end{align}\]
When we are asked to solve for an unknown variable in an equation it means we have to find the value or values of the unknown variable for which the equation satisfies. We are given the following linear equation
\[13=\dfrac{w-14}{2}\]
We see that the unknown variable is here $w$ and we have to find its value. So we eliminate the constants $14,2$ in the right hand side of the given equation. We multiply $2$ both sides of the given equation to have
\[\begin{align}
& \Rightarrow 13\times 2=\dfrac{w-14}{2}\times 2 \\
& \Rightarrow 26=w-14 \\
\end{align}\]
We add 14 both sides of above equation to have;
\[\begin{align}
& \Rightarrow 26+14=w-14+14 \\
& \Rightarrow 40=w \\
\end{align}\]
So the solution of the given equation is $w=40$. \[\]
Note: We note that we can alternatively proceed using the cross-multiplication that is $\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow ad=bc$. So we have $\dfrac{13}{1}=\dfrac{w-14}{2}\Rightarrow 13\times 2=1\times \left( w-14 \right)\Rightarrow 26=w-14$ . We can add 14 as done above to get the solution$w=40$. We should remember that alphabets at the end $u,v,w,x,y,z$ are used to represent variables and alphabets $a,b,c,d,k,l,m,n$are used to represent constants.
Complete step-by-step answer:
We know from algebra that the linear equation in one variable $x$ and constants $a\ne 0,b$is given by
\[ax=b\]
The term with which the variable is multiplied is called variable term and with whom not multiplied is called a constant term. We also know that if we add, subtract, multiply or divide the same number on both sides of the equation, equality holds. It is called balancing the equation. It means for some term $c$ we have,
\[\begin{align}
& ax+c=b+c \\
& ax-c=b-c \\
& ax\times c=b\times c \\
& \dfrac{ax}{c}=\dfrac{b}{c} \\
\end{align}\]
When we are asked to solve for an unknown variable in an equation it means we have to find the value or values of the unknown variable for which the equation satisfies. We are given the following linear equation
\[13=\dfrac{w-14}{2}\]
We see that the unknown variable is here $w$ and we have to find its value. So we eliminate the constants $14,2$ in the right hand side of the given equation. We multiply $2$ both sides of the given equation to have
\[\begin{align}
& \Rightarrow 13\times 2=\dfrac{w-14}{2}\times 2 \\
& \Rightarrow 26=w-14 \\
\end{align}\]
We add 14 both sides of above equation to have;
\[\begin{align}
& \Rightarrow 26+14=w-14+14 \\
& \Rightarrow 40=w \\
\end{align}\]
So the solution of the given equation is $w=40$. \[\]
Note: We note that we can alternatively proceed using the cross-multiplication that is $\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow ad=bc$. So we have $\dfrac{13}{1}=\dfrac{w-14}{2}\Rightarrow 13\times 2=1\times \left( w-14 \right)\Rightarrow 26=w-14$ . We can add 14 as done above to get the solution$w=40$. We should remember that alphabets at the end $u,v,w,x,y,z$ are used to represent variables and alphabets $a,b,c,d,k,l,m,n$are used to represent constants.
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