To determine if (0, 3) is a solution to the system of equations, substitute the values of x and y into the equations and check if the equations are satisfied.
For equation 1:
x + 3y = 9
0 + 3(3) = 9
9 = 9
For equation 2:
2x + 6y = 30
2(0) + 6(3) = 30
18 = 30
Since the second equation is not satisfied, (0, 3) is not a solution to the system of equations.
To solve the system of equations using the elimination method, multiply the first equation by -2 and the second equation by 1/2 to create opposite coefficients for the x-term. The modified equations are:
-2(x + 3y) = -2(9)
1/2(2x + 6y) = 1/2(30)
Simplifying these equations gives:
-2x - 6y = -18
x + 3y = 15
Now, add the two equations together to eliminate the x-term:
(-2x - 6y) + (x + 3y) = -18 + 15
-2x + x - 6y + 3y = -3
-x - 3y = -3
Next, solve this equation for x:
-x = 3y - 3
x = -3y + 3
Substitute this expression for x into one of the original equations to solve for y:
2x + 6y = 30
2(-3y + 3) + 6y = 30
-6y + 6 + 6y = 30
6 = 30
Since 6 does not equal 30, there is no solution to the system of equations.
x + 3y = 9
2x + 6y = 30
A. Is (0, 3) a solution to the system of equations? Explain
B. Solve the system of equations using the elimination method
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