Solve the system by either substitution or elimination
3x-5y=21
2x+y=1
A (2,-3)
B (2,3)
C (-2,-1)
D (-2,1)
a (2, -3)
thanks you for helpful explanation
Well, solving systems of equations is serious business, but let's add a little fun to it!
Let's start by using the elimination method:
We have the two equations:
1) 3x - 5y = 21
2) 2x + y = 1
Let's multiply the second equation by 5 to make the coefficients of y match:
5(2x + y) = 5(1)
10x + 5y = 5
Now, let's add the two equations together:
(3x - 5y) + (10x + 5y) = 21 + 5
13x = 26
x = 2
Now, let's substitute the value of x into one of the original equations:
3(2) - 5y = 21
6 - 5y = 21
-5y = 21 - 6
-5y = 15
y = -3
So, the solution to the system is (x, y) = (2, -3).
Therefore, the correct answer is A (2, -3).
Remember, math can be fun too!
To solve the system of equations, you can use either substitution or elimination method. Let's use the elimination method to solve this system.
Step 1: Multiply the second equation by 5 to make the coefficients of y in both equations opposite:
3x - 5y = 21
2x + y = 1
Multiply the second equation by 5:
10x + 5y = 5
Now we have the system:
3x - 5y = 21
10x + 5y = 5
Step 2: Add the equations together to eliminate y:
(3x - 5y) + (10x + 5y) = 21 + 5
Combine like terms:
13x = 26
Step 3: Divide both sides of the equation by 13 to solve for x:
x = 2
Step 4: Substitute the value of x into one of the equations to solve for y. Let's use the second equation:
2(2) + y = 1
Simplify:
4 + y = 1
Subtract 4 from both sides:
y = -3
Therefore, the solution to the system of equations is (2, -3), which corresponds to option A.
multiply the 2nd equation by 5 and you have
3x-5y = 21
10x+5y = 5
Now you can eliminate y and get x, then figure y.