To solve the system of equations using elimination, we first have the following two equations:
1) [tex]\(5x + 5y = 20\)[/tex]
2) [tex]\(4x + y = 10\)[/tex]
Our goal is to eliminate one of the variables by combining the equations. First, we can simplify the first equation by dividing every term by the common factor of 5:
[tex]\(\frac{5x}{5} + \frac{5y}{5} = \frac{20}{5}\)[/tex]
which simplifies to:
[tex]\(x + y = 4\)[/tex]
Now we have a simplified first equation:
1') [tex]\(x + y = 4\)[/tex]
We can keep the second equation as it is:
2) [tex]\(4x + y = 10\)[/tex]
Now, we want to eliminate one of the variables. Notice that if we multiply the first equation (1') by 4, we will have the same coefficient for [tex]\(x\)[/tex] in both equations:
[tex]\(4(x + y) = 4(4)\)[/tex]
which results in:
[tex]\(4x + 4y = 16\)[/tex]
Now, we have a new system of equations:
3) [tex]\(4x + 4y = 16\)[/tex]
2) [tex]\(4x + y = 10\)[/tex]
To use elimination, we can subtract the second equation from the first one:
[tex]\((4x + 4y) - (4x + y) = 16 - 10\)[/tex]
Simplifying, we get:
[tex]\(4y - y = 6\)[/tex]
Which reduces to:
[tex]\(3y = 6\)[/tex]
Now, to solve for [tex]\(y\)[/tex], we divide both sides by 3:
[tex]\(y = \frac{6}{3}\)[/tex]
This simplifies to:
[tex]\(y = 2\)[/tex]
Now that we have the value of [tex]\(y\)[/tex], we substitute it back into one of the original equations to find [tex]\(x\)[/tex]. We can use the simpler form of the first equation for this:
[tex]\(x + y = 4\)[/tex]
Substituting [tex]\(y = 2\)[/tex] we get:
[tex]\(x + 2 = 4\)[/tex]
Subtract 2 from both sides to solve for [tex]\(x\)[/tex]:
[tex]\(x = 4 - 2\)[/tex]
[tex]\(x = 2\)[/tex]
So the solution to the system of equations is [tex]\(x = 2\)[/tex] and [tex]\(y = 2\)[/tex], which corresponds to the correct answer among the given options.
