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8 Answers

5x+2y=10_3x+2y=6
Since 2y does not contain the variable to solve for, move it to the righthand 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(y5))/(5)_3x+2y=6
Since the equation was solved for x, replace all occurrences of x in the other equations with the solution ((2(y5))/(5)).
x=(2(y5))/(5)_3((2(y5))/(5))+2y=6
Multiply 2 by each term inside the parentheses.
x=((2y+10))/(5)_3((2(y5))/(5))+2y=6
Remove the parentheses around the expression 2y+10.
x=(2y+10)/(5)_3((2(y5))/(5))+2y=6
Divide each term in the numerator by the denominator.
x=(2y)/(5)+(10)/(5)_3((2(y5))/(5))+2y=6
Simplify each expression in (2y)/(5)+(10)/(5).
x=(2y)/(5)+2_3((2(y5))/(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
+(3x)+(2y)=(6) add them up
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
3x2y=6
2x=4
x=2
y=0

x = 2 & y=0..........

x = y