Answer :
Just use elimination
Steps:
1. Multiply everything in the system 7x-y=3 by 2
2. New systems are
3x + 2y=11
14x-2y=6
3. 2y + -2y= 0 so y cancels itself out
4. 14x+ 3x=17x
11 + 6=17
5. New problem is
17x=17
6. Divide each side by 17
7. Your answer is x=1
8. Now substitute the x in 7x-y=3 for 1
7(1)-y=3
7-y=3
9. Subtract 7 on both sides
-y=-4
10. Divide by -1
y=4
Steps:
1. Multiply everything in the system 7x-y=3 by 2
2. New systems are
3x + 2y=11
14x-2y=6
3. 2y + -2y= 0 so y cancels itself out
4. 14x+ 3x=17x
11 + 6=17
5. New problem is
17x=17
6. Divide each side by 17
7. Your answer is x=1
8. Now substitute the x in 7x-y=3 for 1
7(1)-y=3
7-y=3
9. Subtract 7 on both sides
-y=-4
10. Divide by -1
y=4
[tex]\left\{\begin{array}{ccc}3x+2y=11\\7x-y=3&|multiply\ both\ sides\ by\ 2\end{array}\right\\+\underline{\left\{\begin{array}{ccc}3x+2y=11\\14x-2y=6\end{array}\right}\ \ \ |add\ both\ sides\ of\ equations\\. \ \ \ \ \ \ 17x=17\ \ \ \ |divide\ both\ sides\ by\ 17\\.\ \ \ \ \ \ \ \boxed{x=1}\\\\Put\ the\ value\ of\ "x"\ to\ the\ first\ equation:\\\\3(1)+2y=11\\3+2y=11\ \ \ \ |subtract\ 3\ from\ both\ sides\\2y=8\ \ \ \ \ |divide\ both\ sides\ by\ 2\\\boxed{y=4}[/tex]
[tex]Answer:\boxed{\boxed{ \left\{\begin{array}{ccc}x=1\\y=4\end{array}\right}}[/tex]
[tex]Answer:\boxed{\boxed{ \left\{\begin{array}{ccc}x=1\\y=4\end{array}\right}}[/tex]
