Answered

Westonci.ca is the premier destination for reliable answers to your questions, provided by a community of experts. Discover a wealth of knowledge from professionals across various disciplines on our user-friendly Q&A platform. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.

Solving systems algebraically :
4x+2y=7
Y=5x

Sagot :

iceyinyeung
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.
[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]
Thanks for stopping by. We are committed to providing the best answers for all your questions. See you again soon. We hope this was helpful. Please come back whenever you need more information or answers to your queries. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.