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

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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 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

    +(-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

    -3x-2y=-6

    2x=4

    x=2

    y=0

  • x = 2 & y=0..........

  • x = y

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