Answer :
Solving linear systems is sort of like a puzzle, and there are several ways to do this.
One technique is first to try to knock out one variable. You're allowed to manipulate with any of the two, as long as it is the whole equation, and you're allowed to add or subtract one equation from another.
What I would do first if multiply the second equation by 2. That gets me 2y=10x. Now I have 2y in both equations, so if I subtract them, the y variables go away. subtract (4x+2y=7) - (2y=10x), and I get (4x=7-10x). Solve for x: 14x=7 ==> x=0.5.
Now just plug in x to either equation. I choose the second: y=5(0.5) = 2.5.
Then make sure to check your work by plugging in x and y in both equations.
(1) 4x+2y=7 ==> 4(0.5)+2(2.5)=7 ==> 2+5=7 Check
(2) y=5x ==> 2.5=5(2.5) Check
So x=0.5 and y=2.5 solves the system.
One technique is first to try to knock out one variable. You're allowed to manipulate with any of the two, as long as it is the whole equation, and you're allowed to add or subtract one equation from another.
What I would do first if multiply the second equation by 2. That gets me 2y=10x. Now I have 2y in both equations, so if I subtract them, the y variables go away. subtract (4x+2y=7) - (2y=10x), and I get (4x=7-10x). Solve for x: 14x=7 ==> x=0.5.
Now just plug in x to either equation. I choose the second: y=5(0.5) = 2.5.
Then make sure to check your work by plugging in x and y in both equations.
(1) 4x+2y=7 ==> 4(0.5)+2(2.5)=7 ==> 2+5=7 Check
(2) y=5x ==> 2.5=5(2.5) Check
So x=0.5 and y=2.5 solves the system.
[tex] \left\{\begin{array}{ccc}4x+2y=7\\y=5x\end{array}\right\\\\subtitute:\\\\4x+2(5x)=7\\\\4x+10x=7\\\\14x=7\ \ \ \ |divide\ both\ sides\ by\ 14\\\\x=\frac{7}{14}\\\\\boxed{x=\frac{1}{2}}\\\\y=5\times\frac{1}{2}\\\\\boxed{y=\frac{5}{2}}[/tex]
