How do you write $y=\dfrac{3}{4}x+1$ in standard form? \[\]
Answer
631.2k+ views
Hint: We recall that the slope intercept form is $y=mx+c$ and the standard from the linear equation $ax+by=c$ where $a,b,c$ are integers. So we multiply both sides of the equation by and simplify to convert into the form $ax+by=c$.\[\]
Complete step-by-step solution:
We know that equation is a mathematical statement which involves equality between two algebraic expressions. The algebraic expressions contain unknowns called variables like $x,y,z$ and known constants. The highest power on any variable is called its degree.\[\]
We know that a linear equation is an equation with degree 1 which means A linear equation with one variable is given by $ax+b=0$a and in two variables is given by $ax+by+c=0$ where $a,b,c$ are real numbers . The standard form of the linear equation is
\[ax+by=c\]
Here $a,b,c$ have to be integers. We know from the Cartesian coordinate system that every linear equation can be represented as a line. If the line is inclined with positive $x-$axis at an angle $\theta $ then its slope is given by $m=\tan \theta $ and of it cuts $y-$axis at a distance $c$ from the origin the intercept is given by $c$. The slope-intercept form of equation is given by
\[y=mx+c\]
We are given the following linear equation in the question.
\[y=\dfrac{3}{4}x+1\]
We see that the equation is in-slope intercept form. Let us multiply each side of the above equation by 4 to have
\[\begin{align}
& 4\times y=4\times \dfrac{3}{4}\times x+4\times 1 \\
& \Rightarrow 4y=3x+1 \\
\end{align}\]
We subtract both sides above equation by $3x$ to have;
\[\begin{align}
& \Rightarrow 4y-3x=3x-3x+1 \\
& \Rightarrow 4y-3x=1 \\
& \Rightarrow -3x+4y=1 \\
\end{align}\]
So the above equation is in standard form $ax+by=c$ with $a=-3,b=4,c=1$. We can multiply $-1$ both sides to further have
\[\begin{align}
& \Rightarrow -1\left( -3x+4y \right)=-1\times 1 \\
& \Rightarrow 3x-4y=-1 \\
\end{align}\]
So the above equation is in standard form $ax+by=c$ with $a=3,b=-4,c=-1$. \[\]
Note: We note that a linear equation in standard form $ax+by+c=0$ can be converted into slope-intercept form $y=mx+{{c}^{'}}$ using $m=\dfrac{-a}{b}$ and ${{c}^{'}}=\dfrac{c}{b}$. We need at least two linear equations to solve linear equations in two variables. We solve it by the process of elimination or substitution. If there is a unique solution to the equations it will be a point of intersections of the two lines representing the equations.
Complete step-by-step solution:
We know that equation is a mathematical statement which involves equality between two algebraic expressions. The algebraic expressions contain unknowns called variables like $x,y,z$ and known constants. The highest power on any variable is called its degree.\[\]
We know that a linear equation is an equation with degree 1 which means A linear equation with one variable is given by $ax+b=0$a and in two variables is given by $ax+by+c=0$ where $a,b,c$ are real numbers . The standard form of the linear equation is
\[ax+by=c\]
Here $a,b,c$ have to be integers. We know from the Cartesian coordinate system that every linear equation can be represented as a line. If the line is inclined with positive $x-$axis at an angle $\theta $ then its slope is given by $m=\tan \theta $ and of it cuts $y-$axis at a distance $c$ from the origin the intercept is given by $c$. The slope-intercept form of equation is given by
\[y=mx+c\]
We are given the following linear equation in the question.
\[y=\dfrac{3}{4}x+1\]
We see that the equation is in-slope intercept form. Let us multiply each side of the above equation by 4 to have
\[\begin{align}
& 4\times y=4\times \dfrac{3}{4}\times x+4\times 1 \\
& \Rightarrow 4y=3x+1 \\
\end{align}\]
We subtract both sides above equation by $3x$ to have;
\[\begin{align}
& \Rightarrow 4y-3x=3x-3x+1 \\
& \Rightarrow 4y-3x=1 \\
& \Rightarrow -3x+4y=1 \\
\end{align}\]
So the above equation is in standard form $ax+by=c$ with $a=-3,b=4,c=1$. We can multiply $-1$ both sides to further have
\[\begin{align}
& \Rightarrow -1\left( -3x+4y \right)=-1\times 1 \\
& \Rightarrow 3x-4y=-1 \\
\end{align}\]
So the above equation is in standard form $ax+by=c$ with $a=3,b=-4,c=-1$. \[\]
Note: We note that a linear equation in standard form $ax+by+c=0$ can be converted into slope-intercept form $y=mx+{{c}^{'}}$ using $m=\dfrac{-a}{b}$ and ${{c}^{'}}=\dfrac{c}{b}$. We need at least two linear equations to solve linear equations in two variables. We solve it by the process of elimination or substitution. If there is a unique solution to the equations it will be a point of intersections of the two lines representing the equations.
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