Question

How do you solve $5\left( x-4 \right)=4\left( x+1 \right)$ ? \[\]

Answer 1

Hint: We open the bracket in both sides and use distributive law of multiplication over s and subtraction and addition to simplify. We add and subtract suitable values so that we can collect variable terms at one side of the equation and the constant terms at the other side. We divide the coefficient of the variable $x$ to find the value of the variable.

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 constant term. The constant that is multiplied with the variable $x$ is called the coefficient of that variable which here is $a$. 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,
\[\begin{align}
  & ax+c=b+c \\
 & ax-c=b-c \\
 & ax\times c=b\times c \\
 & \dfrac{ax}{c}=\dfrac{b}{c}\left( c\ne 0 \right) \\
\end{align}\]
 When we are asked to solve an equation we find the values of unknown variables $x$. If we are given variable terms on both sides for example $ax+b=cx+d$ then we collect the like terms (variable terms and constant terms) at two different sides. We are asked to solve the following equation
\[ 5\left( x-4 \right)=4\left( x+1 \right)\]
We see that the variable $x$ is at both sides of the equation. So we first open the bracket using use distributive law of multiplication over subtraction that is $a\left( b-c \right)=a\times b-a\times c$ and over addition $a\left( b+c \right)=a\times b+a+c$. We proceed to have
\[\begin{align}
  & \Rightarrow 5\times x-5\times 4=4\times x+4\times 1 \\
 & \Rightarrow 5x-20=4x+4 \\
\end{align}\]
 We subtract both sides by $4x$ in the above step to have-
\[\begin{align}
  & \Rightarrow 5x-4x-20=4x-4x+4 \\
 & \Rightarrow x-20=4 \\
\end{align}\]
We add both sides 20 in the above step to have-
\[\begin{align}
  & \Rightarrow x-20+20=20+4 \\
 & \Rightarrow x=24 \\
\end{align}\]
So the solution of the given equation is $x=24$

Note:
We note that since the coefficient of $x$ is 1 in the equation $x=1$ we do not need to divide it further. We can put the solution $x=24$ in the left hand side expression of the given equation to have $5\left( x-4 \right)=5\left( 24-4 \right)=5\times 20=100$ and we put $x=24$in right hand side expression to have$4\left( x+1 \right)=4\left( 24+1 \right)=4\times 25=100$. Since values at left hand side and right side are equal our solution is verified.