Question

iii) Which of the following is a solution of the equation
2x+3y-13.
1
​

Answer 1

Answer:

Explanation:

⇒2x+3y=13−−−−(1)

And 3x−5y=−9−−−−(2)

Now multiply 3 to the first equation and multiply 2 to the second equation, we get,

⇒2x+3y=13−−−−(1)×3

     3x−5y=−9−−−−(2)×2,

Now simplifying we get,

⇒6x+9y=39−−−−(1)

     6x−10y=−18−−−−(2)

Now subtracting second equation from the first equation we get,

⇒6x+9y=39−−−−(1)

6x−10y=−18−−−−(2)

(−) (−) (−)

Now simplifying we get,

⇒9y−(−10y)=39−(−18),

Now simplifying we get,

⇒9y+10y=39+18,

Now adding the terms we get,

⇒19y=57,

Now divide both sides of the equation with 19 we get,

⇒19y19=5719,

Now simplifying we get,

⇒y=3,

Now substituting the value ofyin the first equation, we get,

⇒2x+3(3)=13,

Now multiplying we get,

⇒2x+9=13,

Now subtract 9 from both sides of the equation we get,

⇒2x+9−9=13−9,

Now simplifying we get,

⇒2x=4,

Now divide both sides of the equation with 2 we get,

⇒2x2=42,

Now simplifying we get,

⇒x=2,

So, the values of x and y are 2 and 3.

∴ The value of x and y when the system of equations 2x+3y=13 and 3x−5y=−9are solved are 2 and 3.

Note:

The three methods which are most commonly used to solve systems of equations are substitution, elimination and augmented matrices.

Substitution and elimination are simpler methods of solving equations and are used much more frequently than augmented matrices in basic algebra. The substitution method is especially useful when one of the variables is already isolated in one of the equations. The elimination method is useful when the coefficient of one of the variables is the same (or its negative equivalent) in all of the equations. The primary advantage of augmented matrices is that it can be used to solve systems of three or more equations in situations where substitution and elimination are either unfeasible or impossible.⇒2x+3y=13−−−−(1)

And 3x−5y=−9−−−−(2)

Now multiply 3 to the first equation and multiply 2 to the second equation, we get,

⇒2x+3y=13−−−−(1)×3

     3x−5y=−9−−−−(2)×2,

Now simplifying we get,

⇒6x+9y=39−−−−(1)

     6x−10y=−18−−−−(2)

Now subtracting second equation from the first equation we get,

⇒6x+9y=39−−−−(1)

6x−10y=−18−−−−(2)

(−) (−) (−)

Now simplifying we get,

⇒9y−(−10y)=39−(−18),

Now simplifying we get,

⇒9y+10y=39+18,

Now adding the terms we get,

⇒19y=57,

Now divide both sides of the equation with 19 we get,

⇒19y19=5719,

Now simplifying we get,

⇒y=3,

Now substituting the value ofyin the first equation, we get,

⇒2x+3(3)=13,

Now multiplying we get,

⇒2x+9=13,

Now subtract 9 from both sides of the equation we get,

⇒2x+9−9=13−9,

Now simplifying we get,

⇒2x=4,

Now divide both sides of the equation with 2 we get,

⇒2x2=42,

Now simplifying we get,

⇒x=2,

So, the values of x and y are 2 and 3.

∴ The value of x and y when the system of equations 2x+3y=13 and 3x−5y=−9are solved are 2 and 3.

Note:

The three methods which are most commonly used to solve systems of equations are substitution, elimination and augmented matrices.

Substitution and elimination are simpler methods of solving equations and are used much more frequently than augmented matrices in basic algebra. The substitution method is especially useful when one of the variables is already isolated in one of the equations. The elimination method is useful when the coefficient of one of the variables is the same (or its negative equivalent) in all of the equations. The primary advantage of augmented matrices is that it can be used to solve systems of three or more equations in situations where substitution and elimination are either unfeasible or impossible.⇒2x+3y=13−−−−(1)

And 3x−5y=−9−−−−(2)

Now multiply 3 to the first equation and multiply 2 to the second equation, we get,

⇒2x+3y=13−−−−(1)×3

     3x−5y=−9−−−−(2)×2,

Now simplifying we get,

⇒6x+9y=39−−−−(1)

     6x−10y=−18−−−−(2)

Now subtracting second equation from the first equation we get,

⇒6x+9y=39−−−−(1)

6x−10y=−18−−−−(2)

(−) (−) (−)

Now simplifying we get,

⇒9y−(−10y)=39−(−18),

Now simplifying we get,

⇒9y+10y=39+18,

Now adding the terms we get,

⇒19y=57,

Now divide both sides of the equation with 19 we get,

⇒19y19=5719,

Now simplifying we get,

⇒y=3,

Now substituting the value ofyin the first equation, we get,

⇒2x+3(3)=13,

Now multiplying we get,

⇒2x+9=13,

Now subtract 9 from both sides of the equation we get,

⇒2x+9−9=13−9,

Now simplifying we get,

⇒2x=4,

Now divide both sides of the equation with 2 we get,

⇒2x2=42,

Now simplifying we get,

⇒x=2,

So, the values of x and y are 2 and 3.

∴ The value of x and y when the system of equations 2x+3y=13 and 3x−5y=−9are solved are 2 and 3.

Note:

The three methods which are most commonly used to solve systems of equations are substitution, elimination and augmented matrices.