# Solve the system algebraically. 5x + 2y = 10 3x + 2y = 6?

choose wisely

• 5x+2y=10_3x+2y=6

Since 2y does not contain the variable to solve for, move it to the right-hand side of the equation by subtracting 2y from both sides.

5x=-2y+10_3x+2y=6

Divide each term in the equation by 5.

x=-(2(y-5))/(5)_3x+2y=6

Since the equation was solved for x, replace all occurrences of x in the other equations with the solution (-(2(y-5))/(5)).

x=-(2(y-5))/(5)_3(-(2(y-5))/(5))+2y=6

Multiply -2 by each term inside the parentheses.

x=((-2y+10))/(5)_3(-(2(y-5))/(5))+2y=6

Remove the parentheses around the expression -2y+10.

x=(-2y+10)/(5)_3(-(2(y-5))/(5))+2y=6

Divide each term in the numerator by the denominator.

x=-(2y)/(5)+(10)/(5)_3(-(2(y-5))/(5))+2y=6

Simplify each expression in -(2y)/(5)+(10)/(5).

x=-(2y)/(5)+2_3(-(2(y-5))/(5))+2y=6

Multiply -2 by each term inside the parentheses.

x=-(2y)/(5)+2_3(((-2y+10))/(5))+2y=6

Remove the parentheses around the expression -2y+10.

x=-(2y)/(5)+2_3((-2y+10)/(5))+2y=6

Divide each term in the numerator by the denominator.

x=-(2y)/(5)+2_3(-(2y)/(5)+(10)/(5))+2y=6

Simplify each expression in -(2y)/(5)+(10)/(5).

x=-(2y)/(5)+2_3(-(2y)/(5)+2)+2y=6

Multiply 3 by each term inside the parentheses.

x=-(2y)/(5)+2_(-(6y)/(5)+6)+2y=6

Reorder the polynomial -(6y)/(5)+6+2y alphabetically from left to right, starting with the highest order term.

x=-(2y)/(5)+2_-(6y)/(5)+2y+6=6

Find the LCD (least common denominator) of -(6y)/(5)+2y+0.

x=-(2y)/(5)+2_Least common denominator: 5

Multiply each term in the equation by 5 in order to remove all the denominators from the equation.

x=-(2y)/(5)+2_4y=0

Since the equation was solved for y, replace all occurrences of y in the other equations with the solution (0).

x=-(2y)/(5)+2_y=0

Since the equation was solved for y, replace all occurrences of y in the other equations with the solution (0).

x=-(2(0))/(5)+2_y=0

Multiply -2*0 to get 0 in the numerator.

x=(0)/(5)+2_y=0

Any expression with a factor of 0 in the numerator is equal to 0.

x=0+2_y=0

Remove the 0 from the polynomial; adding or subtracting 0 does not change the value of the expression.

x=2_y=0

This is the solution to the system of equations.

x=2_y=0

• 5x+2y=10

(3x+2y=6) time this by -1

5x+2y=10

2x=4 divide it by 2

x=2

now plug in the 2 for x to solve the y

5(2)+2y=10

10+2y=10 minus 10

2y=0

y=0

• Solve The System Algebraically

• x=2

y=0

You can solve this as a system of equations in the form Ax=b and find x with A^-1*b

A = [5 2; 3 2] (the coefficients of x and y)

b = [10; 6] (the results)

• 5x + 2y = 10.........(i)

3x + 2y = 6..........(ii)

Subtract (ii) from (i), we get

(5x + 2y) −( 3x + 2y)= 10−6

or 2x = 4

hence x= 2

From (i) plugging the value of x, we get

5×2 + 2y = 10

10+ 2y = 10

2y=10−=0

Hence y= 0

• 5x+2y=10

3x+2y=6

5x+2y=10

-1(3x+2y=6)

5x+2y=10

-3x-2y=-6

2x=4

x=2

y=0

• x = 2 & y=0..........

Source(s): i passed elementary school long back.........
• x = y