For this case we have a system of two linear equations with two igcognitas, given by:
[tex]5v + 4w = 1\\3v-6w = 2[/tex]
Where v and w are the unknowable variables.
To solve, we perform the following steps:
1st step:
We multiply the first equation by 3:
[tex]15v + 12w = 3[/tex]
2nd step:
We multiply the second equation by -5:
[tex]-15v + 30w = -10[/tex]
3rd step:
We add the equations:
[tex]15v + 12w = 3\\-15v + 30w = -10[/tex]
We have then:
[tex]42w = -7[/tex]
[tex]w =-\frac{7}{42}[/tex]
[tex]w =-\frac{1}{6}[/tex]
Thus, the value of w is [tex]-\frac{1}{6}[/tex].
4th step:
We substitute [tex]w =-\frac{1}{6}[/tex] in any of the equations:
[tex]3v-6w = 2\\3v-6 * (-\frac{1}{6}) = 2[/tex]
[tex]3v + 1 = 2\\3v = 2-1\\3v = 1[/tex]
[tex]v =\frac{1}{3}[/tex]
So, the value of v is [tex]\frac{1}{3}[/tex]
Answer:
The value of w is [tex]-\frac{1}{6}[/tex]
