Answer:
(1, 1)
Step-by-step explanation:
Solve the System of Equations
5x − 2y = 3
2x + 5y = 7
Solve for x in the first equation.
x = 3/5 + 2y
/5
2x + 5y = 7
Replace all occurrences of x with
3/5 +2y
/5 in each equation.
Replace all occurrences of x in 2x
+ 5y
= 7 with
3/5 + 2y/5
2 ( (3/5) + (2y/5))+ 5y = 7
x = 3/5 + 2y
/5
Simplify 2((3/5)+(2y/5))
+5
6 + 29y = 7 ⋅(5)
x = (3/5)+(2y/5)
Solve for y in the first equation.
Multiply both sides of the equation by 5.
6 + 29y = 7 ⋅ (5)
x = (3/5)+(2y/5)
Remove parentheses.
6 + 29y = 7 ⋅ (5)
x = (3/5)+(2y/5)
Multiply 7 by 5.
6 + 29y = 35
x = (3/5)+(2y/5)
Move all terms not containing y to the right side of the equation.
Subtract 6 from both sides of the equation.
29y = 35 − 6
x = (3/5)+(2y/5)
Subtract 6 from 35.
29y = 29
x = (3/5)+(2y/5)
Divide each term by 29 and simplify.
Divide each term in 29y = 29 by 29.
29y/29 = 29/29
x = (3/5)+(2y/5)
Cancel the common factor of 29.
y = 29
/29
x = (3/5)+(2y/5)
Divide 29 by 29.
y = 1
x = (3/5)+(2y/5)
Replace all occurrences of y with 1 in each equation.
Replace all occurrences of y in x = (3/5)+(2y/5) with 1.
x = (3/5)+(2(1)/5)3
y = 1
Simplify
x = (3/5)+(2(1)/5)3
y = 1
x = 1
y = 1
The solution to the system is the complete set of ordered pairs that are valid solutions.
(1, 1)
The result can be shown in multiple forms.
Point Form:
(1, 1)
Equation Form:
x = 1, y = 1
