Question

How do you solve \[5x - y = 9\] and \[x = \dfrac{1}{2}y - 3\] ?

Answer 1

Hint: To solve the given simultaneous equation, substitute the given equation for \[x\] of equation 2 in equation 1 and hence simplify the terms by substituting the equation to get the value of \[y\] . Substitute the value obtained of \[y\]in equation 2, hence the value of \[x\] is obtained.

Complete step-by-step solution:
Let us write the given equation
\[5x - y = 9\] …………………………. 1
\[x = \dfrac{1}{2}y - 3\] ..………………………… 2
The given equation 1 is standard equation
\[Ax + By = C\]
Substitute the given value of \[x\], of equation 2 in equation 1
\[5x - y = 9\]
\[5\left( {\dfrac{1}{2}y - 3} \right) - y = 9\]
Simplify the equation with respect to y term we get
\[\dfrac{5}{2}y - 15 - y = 9\]
Now add 15 on both sides of the equation
\[\dfrac{5}{2}y - 15 - y + 15 = 9 + 15\]
\[\dfrac{5}{2}y - y = 24\]
In the obtained equation we can observe that -15 and +15 implies to zero. Hence, we get the equation as:
\[\dfrac{5}{2}y - \dfrac{y}{1} = 24\]
Simplifying the above equation, we get
\[\dfrac{{5y - 2y}}{2} = 24\]
\[\dfrac{{3y}}{2} = 24\]
Here we need to find the value of \[y\], hence we get
\[3y = 24\left( 2 \right)\]
\[3y = 48\]
Simplifying the terms, to get the value of \[y\]as:
\[y = \dfrac{{48}}{3}\]
Therefore, the value of \[y\]is
\[y = 16\]
As we got the value of \[y\] as 16, now substitute the value of \[y\]in equation 2 i.e.,
\[x = \dfrac{1}{2}y - 3\]
\[x = \dfrac{1}{2}\left( {16} \right) - 3\]
Simplifying the terms of the equation we get,
\[x = 8 - 3\]
\[x = 5\]
Therefore, the values obtained of \[x\] and \[y\]is
\[x = 5\] and \[y = 16\].

Note: We know that Simultaneous equations are two equations, each with the same two unknowns and are "simultaneous" because they are solved together, hence the key point to solve this kind of equations is we need to combine all the terms and then simplify the terms to get the value of \[x\] also the value of \[y\].