How do you solve $20=5\left( -3+x \right)$ ? \[\]
Answer
585k+ views
Hint: We recall the linear equations and basic properties of linear equations. We solve the given linear equation by opening the parenthesis in the right hand and then eliminating the constants at the right hand side till the variable is at one side and constant at the other side. We finally divide both sides by a suitable number to get only $x$ at one side.
Complete step-by-step solution:
We know from algebra that the linear equation in one variable $x$ and constants $a,b$ is given by
\[ax=b\]
Here $a$ and $b$ are real numbers called constant of the equation where $a$ cannot be zero. The side left to the sign of equality ${}^{'}{{=}^{'}}$ is called the left hand side and right hand side of the equation. We also know that if we add, subtract, multiply or divide the same number on both sides of the equation, equality holds. It means for some real number $c$ we have,
\[\begin{align}
& ax+c=b+c \\
& ax-c=b-c \\
& ax\times x=b\times c \\
& \dfrac{ax}{c}=\dfrac{b}{c} \\
\end{align}\]
When we are asked to solve for $x$ in an equation it means we have to find the value or values of $x$ for which the equation satisfies. We are given the following linear equation
\[20=5\left( -3+x \right)\]
We open the bracket in the right hand side using distributive law of multiplication over addition to have
\[\begin{align}
& 20=5\times \left( -3 \right)+5\times x \\
& \Rightarrow 20=-15+5x \\
\end{align}\]
We add 15 both sides of the above equation to have;
\[\begin{align}
& 20+15=-15+5x+15 \\
& \Rightarrow 35=-15+15+5x \\
& \Rightarrow 35=5x \\
\end{align}\]
We divide both sides of the above equation by 7 to have;
\[\begin{align}
& \Rightarrow \dfrac{35}{5}=\dfrac{5x}{5} \\
& \Rightarrow 7=x \\
& \Rightarrow x=7 \\
\end{align}\]
So the value of $x$ is 7.
Note: We can verify that the obtained value $x=7$ is correct by putting in the right hand side of the equation and getting the result at the left hand side that is $5\left( -3+7 \right)=5\times 4=10$. The highest power on the variable is called degree and if degree is 1 we call the equation linear equation. The linear equation of two variables is given by $ax+by+c=0$ where $a,b$ are non-zero constants.
Complete step-by-step solution:
We know from algebra that the linear equation in one variable $x$ and constants $a,b$ is given by
\[ax=b\]
Here $a$ and $b$ are real numbers called constant of the equation where $a$ cannot be zero. The side left to the sign of equality ${}^{'}{{=}^{'}}$ is called the left hand side and right hand side of the equation. We also know that if we add, subtract, multiply or divide the same number on both sides of the equation, equality holds. It means for some real number $c$ we have,
\[\begin{align}
& ax+c=b+c \\
& ax-c=b-c \\
& ax\times x=b\times c \\
& \dfrac{ax}{c}=\dfrac{b}{c} \\
\end{align}\]
When we are asked to solve for $x$ in an equation it means we have to find the value or values of $x$ for which the equation satisfies. We are given the following linear equation
\[20=5\left( -3+x \right)\]
We open the bracket in the right hand side using distributive law of multiplication over addition to have
\[\begin{align}
& 20=5\times \left( -3 \right)+5\times x \\
& \Rightarrow 20=-15+5x \\
\end{align}\]
We add 15 both sides of the above equation to have;
\[\begin{align}
& 20+15=-15+5x+15 \\
& \Rightarrow 35=-15+15+5x \\
& \Rightarrow 35=5x \\
\end{align}\]
We divide both sides of the above equation by 7 to have;
\[\begin{align}
& \Rightarrow \dfrac{35}{5}=\dfrac{5x}{5} \\
& \Rightarrow 7=x \\
& \Rightarrow x=7 \\
\end{align}\]
So the value of $x$ is 7.
Note: We can verify that the obtained value $x=7$ is correct by putting in the right hand side of the equation and getting the result at the left hand side that is $5\left( -3+7 \right)=5\times 4=10$. The highest power on the variable is called degree and if degree is 1 we call the equation linear equation. The linear equation of two variables is given by $ax+by+c=0$ where $a,b$ are non-zero constants.
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