Answer:
x = 3 and y = 2 or (3,2) in coordinate form
Step-by-step explanation:
Given:
[tex]\displaystyle \large{\begin{cases} x-y=1 \\ x+3y=9 \end{cases}}[/tex]
Solve by elimination of x-terms by multiplying either one of equations with -1:
For this, I choose to multiply -1 in first equation:
[tex]\displaystyle \large{\begin{cases} -1(x-y=1) \\ x+3y=9 \end{cases}}\\\displaystyle \large{\begin{cases} -x+y=-1 \\ x+3y=9 \end{cases}}[/tex]
Then add both equations:
[tex]\displaystyle \large{-x+x+y+3y=-1+9}\\\displaystyle \large{4y=8}[/tex]
Divide both sides:
[tex]\displaystyle \large{\dfrac{4y}{4} = \dfrac{8}{4}}\\\displaystyle \large{y=2}[/tex]
Next, substitute y = 2 in one of equations to solve for x:
For this, I choose to substitute y = 2 in the first equation:
[tex]\displaystyle \large{x-y=1}[/tex]
Substitute y = 2 in:
[tex]\displaystyle \large{x-2=1}[/tex]
Add both sides by 2:
[tex]\displaystyle \large{x-2+2=1+2}\\\displaystyle \large{x=3}[/tex]
Hence, the solution is x = 3 and y = 2 or you can write as in coordinate form (3,2)
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First method is elimination method, I’ll demonstrate another method which is by using matrices to find the solutions.
Given:
[tex]\displaystyle \large{\begin{cases} x-y=1 \\ x+3y=9 \end{cases}}[/tex]
Write the system of equations in matrices form:
[tex]\displaystyle \large{\left[\begin{array}{ccc}1&-1\\1&3\end{array}\right] \left[\begin{array}{ccc}x\\y\end{array}\right] =\left[\begin{array}{ccc}1\\9\end{array}\right]}[/tex]
Let:
[tex]\displaystyle \large{A=\left[\begin{array}{ccc}1&-1\\1&3\end{array}\right] }\\\displaystyle \large{X=\left[\begin{array}{ccc}x\\y\end{array}\right] \\\displaystyle \large{B = \left[\begin{array}{ccc}1\\9\end{array}\right] }[/tex]
From [tex]\displaystyle \large{AX=B \to X=A^{-1}B}[/tex] where:
[tex]\displaystyle \large{A^{-1}=\dfrac{1}{\det A} \left[\begin{array}{ccc}d&-b\\-c&a\end{array}\right] = \dfrac{1}{ad-bc} \left[\begin{array}{ccc}d&-b\\-c&a\end{array}\right] }[/tex]
Find [tex]\displaystyle \large{\det A}[/tex]:
[tex]\displaystyle \large{\det A = ad-bc = 3+1=4}[/tex]
Therefore:
[tex]\displaystyle \large{A^{-1} = \dfrac{1}{4}\left[\begin{array}{ccc}3&1\\-1&1\end{array}\right]}\\\therefore \displaystyle \large{X = \dfrac{1}{4}\left[\begin{array}{ccc}3&1\\-1&1\end{array}\right] \left[\begin{array}{ccc}1\\9\end{array}\right]}[/tex]
Evaluate the matrices:
[tex]\displaystyle \large{X= \dfrac{1}{4}\left[\begin{array}{ccc}3(1)+1(9)\\-1(1)+1(9)\end{array}\right] }\\\displaystyle \large{X= \dfrac{1}{4}\left[\begin{array}{ccc}3+9\\-1+9\end{array}\right] }\\\displaystyle \large{X= \dfrac{1}{4}\left[\begin{array}{ccc}12\\8\end{array}\right]}\\\displaystyle \large{X= \left[\begin{array}{ccc}3\\2\end{array}\right] }\\\therefore \displaystyle \large{\left[\begin{array}{ccc}x\\y\end{array}\right] = \left[\begin{array}{ccc}3\\2\end{array}\right]}[/tex]
Therefore, the solution is x = 3 and y = 2
