Question

Solve each system of equations
5x- 2y = 3
2x + 5y = 7 Ordered pair​

Answer 1

(1, 1)

Explanation:

We can solve each system of equations by using the elimination method for the first variable and then substituting for the second variable.

To eliminate, we must make one the same by multiplying the equation by a constant. Let's make x the same by multiplying the first equation with 2 and the second equation with 5. Both should get 10x.

10x - 4y = 6

10x + 25y = 35

Now we can eliminate x by subtracting these equations (this works since 10x - 10x = 0). So:

-4y - 25y = 6 - 35

-29y = -29

y = 1

Now that we have y, we can substitute it into either equation and solve for x.

5x - 2(1) = 3

5x = 5

x = 1

Answer: (1, 1)

Answer 2

(1, 1)

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