Respuesta :
Answer:
To answer this question one can do either one of two methods one of which is elimination, or substitution:
Step-by-step explanation:
One these methods is two fold and that is one can be done with linear algebra which can represent the system of equations using a augmented matrix.
[tex]4x+y=2\\-x+y=-3[/tex]
Now we can setup the Gauss-elimination using row echelon form:
[tex]\left[\begin{array}{ccc}4&1&|+2\\-1&1&|-3\end{array}\right][/tex]
The first thing we are gonna do is perform the indicated row operation
[tex]\frac{1}4R_1+R_2[/tex]
Which gives you the following:
[tex]\left[\begin{array}{ccc}4&1&|+2\\0&\frac{5}4&|-\frac{5}2\end{array}\right][/tex]
Now we must perform another row operation
[tex]4*R_2[/tex]
[tex]\left[\begin{array}{ccc}4&1&|2\\ 0 &5&|-10\end{array}\right][/tex]
Next row operation is the following:
[tex]R_1-\frac{1}5R_2[/tex]
Which produces the following matrix
[tex]\left[\begin{array}{ccc}4&0&|4\\0&5&|-10\end{array}\right][/tex]
Then chain in these two row operations and you get the following:
[tex]\frac{R_1}4\\\frac{R_2}5[/tex]
Which produces this matrix:
[tex]\left[\begin{array}{ccc}1&0&1\\0&1&-2\end{array}\right][/tex]
Which means the ordered pair is the following:
[tex](1,-2)[/tex] is the solution
