Respuesta :
To solve systems of two equations with two variables (in this case, "x" and "y"), you choose one equation and attempt to solve for one of the variables.
For example, I will try solving for x using the first equation.
5x - 2y = 5
5x = 5 + 2y [move 2y to other side]
x = (5 + 2y) / 5 [divide both sides by 5]
x = 1 + (.4)y [simplified form]
Now, I plug this value into the x of the second equation to solve for y.
2x + 3y = 21
2 * (1 + (.4y)) + 3y = 21 [plug in x]
2 + (.8)y + 3y = 21 [distribute 2 to the x value we plugged in]
2 + 3.8y = 21 [add the y terms together]
3.8y = 19 [subtract 2]
y = 5 [divide by 5]
Now that we know the value of y, plug it into either of the original equations to find x. I'll plug it into the first one.
5x - 2y = 5
5x - 2(5) = 5 [plug in y]
5x - 10 = 5 [2 * 5]
5x = 15 [add 10]
x = 3 [divide by 5]
The final answer is x = 3, y = 5.
For example, I will try solving for x using the first equation.
5x - 2y = 5
5x = 5 + 2y [move 2y to other side]
x = (5 + 2y) / 5 [divide both sides by 5]
x = 1 + (.4)y [simplified form]
Now, I plug this value into the x of the second equation to solve for y.
2x + 3y = 21
2 * (1 + (.4y)) + 3y = 21 [plug in x]
2 + (.8)y + 3y = 21 [distribute 2 to the x value we plugged in]
2 + 3.8y = 21 [add the y terms together]
3.8y = 19 [subtract 2]
y = 5 [divide by 5]
Now that we know the value of y, plug it into either of the original equations to find x. I'll plug it into the first one.
5x - 2y = 5
5x - 2(5) = 5 [plug in y]
5x - 10 = 5 [2 * 5]
5x = 15 [add 10]
x = 3 [divide by 5]
The final answer is x = 3, y = 5.