
How do you solve $y = 7x - 1$ and $y = - x + 14$ using substitution method?
Answer
544.8k+ views
Hint:The substitution method is used to solve simultaneous linear equations. In this method, the value of one variable either $x$ or $y$ from one equation is substituted in the other equation and then solve the equation to get the value of one variable and then put the value of this variable in either of the equations to get the value of the other variable.
Complete step by step answer:
The given two linear equations are
$y = 7x - 1$ - - - - - - - - - - - - - - - - - - - - - - - -(1)
$y = - x + 14$ - - - - - - - - - - - - - - - - - - - - - - (2)
From the first equation we get $y = 7x - 1$ .
Now, put the value of $y$ in the second equation and then we get,
$
\Rightarrow y = - x + 14 \\
\Rightarrow 7x - 1 = - x + 14 \\
\Rightarrow 7x + x = 14 + 1 \\
\Rightarrow 8x = 15 \\
\therefore x = \dfrac{{15}}{8}
$
Now, put the value of variable $x = \dfrac{{15}}{8}$ in the second equation $y = - x + 14$.
$
\Rightarrow y = - x + 14 \\
\Rightarrow y = - \left( {\dfrac{{15}}{8}} \right) + 14 \\
\Rightarrow y = \dfrac{{ - 15 + 112}}{8} \\
\therefore y = \dfrac{{97}}{8}
$
Thus, the value of variables $x$ and $y$ are $\dfrac{{15}}{8}$ and $\dfrac{{97}}{8}$ respectively.
Hence, by substitution method we get $\left( {\dfrac{{15}}{8},\dfrac{{97}}{8}} \right)$ is the required solution of the given two equations.
Note: The given two linear equations are also solved by elimination method. In this method our main aim is to eliminate one variable either $x$ or $y$. For this multiply first equation by some numerals and the second equation by some other numerals such that addition or subtraction of the two equation cancel either the variable $x$ or $y$, then solve the value of one variable and put this value in either of the equation to get the value of the other variable.
Complete step by step answer:
The given two linear equations are
$y = 7x - 1$ - - - - - - - - - - - - - - - - - - - - - - - -(1)
$y = - x + 14$ - - - - - - - - - - - - - - - - - - - - - - (2)
From the first equation we get $y = 7x - 1$ .
Now, put the value of $y$ in the second equation and then we get,
$
\Rightarrow y = - x + 14 \\
\Rightarrow 7x - 1 = - x + 14 \\
\Rightarrow 7x + x = 14 + 1 \\
\Rightarrow 8x = 15 \\
\therefore x = \dfrac{{15}}{8}
$
Now, put the value of variable $x = \dfrac{{15}}{8}$ in the second equation $y = - x + 14$.
$
\Rightarrow y = - x + 14 \\
\Rightarrow y = - \left( {\dfrac{{15}}{8}} \right) + 14 \\
\Rightarrow y = \dfrac{{ - 15 + 112}}{8} \\
\therefore y = \dfrac{{97}}{8}
$
Thus, the value of variables $x$ and $y$ are $\dfrac{{15}}{8}$ and $\dfrac{{97}}{8}$ respectively.
Hence, by substitution method we get $\left( {\dfrac{{15}}{8},\dfrac{{97}}{8}} \right)$ is the required solution of the given two equations.
Note: The given two linear equations are also solved by elimination method. In this method our main aim is to eliminate one variable either $x$ or $y$. For this multiply first equation by some numerals and the second equation by some other numerals such that addition or subtraction of the two equation cancel either the variable $x$ or $y$, then solve the value of one variable and put this value in either of the equation to get the value of the other variable.
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