Question

Solve: $10x-5-7x=5x+15-8$.\[\]

Answer 1

Hint: We solve the given linear equation in one variable by collecting the variable terms at left hand side and the constant terms in the right hand side and simplify by adding or subtracting same term both side until we get a term only $x$ in left hand side. We add 5 both sides, subtract $-5x$ both sides and the variable and constant terms. We get $-2x$in left hand side and we divide the equation by $-2$ to get $x$.\[\]

Complete step by step answer:
We know that linear equation in one variable is given by
\[ax+b=0\]
Here $a,b$ are real numbers called constants and $x$ is called unknown or variable. The equation is called linear because the power on $x$ is 1. The constants that are multiplied with a variable are called coefficients of the said variable. Here the coefficient of $x$ is $a$ which must not be zero. The linear equation in one variable is solved by collecting variable terms and constant terms at different side.\[\]
We are given the equation in question
\[10x-5-7x=5x+15-8\]
We see that the given equation there is only one variable $x$. Let us collect the terms multiplied with $x$ and the constant terms close to each other using the commutative property of addition. We have
\[\Rightarrow 10x-7x-5=5x+15-8\]
Let us add 5 on both sides of the equation and simplify. We have,
\[\begin{align}
  & \Rightarrow 10x-7x-5+5=5x+15-8+5 \\
 & \Rightarrow 10x-7x=5x+7+5 \\
 & \Rightarrow 10x-7x=5x+12 \\
\end{align}\]
Let us subtract $5x$ from both side of the equation to have,
\[\begin{align}
  & \Rightarrow 10x-7x-5x=5x+12-5x \\
 & \Rightarrow 10x-7x-5x=5x-5x+12 \\
\end{align}\]
Let us take $x$ common in both sides of the equation from terms where $x$ is multiplied. We have,
\[\begin{align}
  & \Rightarrow x\left( 10-7-5 \right)=x\left( 5-5 \right)+12 \\
 & \Rightarrow x\left( -2 \right)=x\times 0+12 \\
 & \Rightarrow -2x=12 \\
\end{align}\]
Let us divided both side of equation by $-2$ and then cancel out the terms to have the solution as

\[x=\dfrac{12}{-2}=-6\]

Note: We note that we are asked to solve for $x$, it is asking for what values of $x$ the equation satisfies which are also called solutions of the equation. We can put $x=-6$ in a given equation to verify the satisfaction. The linear equation in two variables is given as $ax+by+c=0$ and we need at least 2 equations to solve for $x,y$ unlike 1 equation for linear equation in one variable.