Solving with the method of equalization for systems of two linear equations with two unknowns - Examples, Exercises and Solutions

To solve systems of two linear equations with two unknowns with the equating method, we must arrive at a situation in which one of the coefficients is equal or opposite in the same unknown in the two equations.

How do we do this?

  • We will arrange the equation in such a way that the types of unknowns lie one on top of the other respectively.
  • We will carry out a multiplication in one or both equations so that in the two equations there is an identical or opposite coefficient in one of the unknowns (Example of opposite coefficient: 4 4 and 4-4 ). The multiplication will be done separately for each term.
  • We will add or subtract the equations as necessary, in this way we will suppress an unknown and we will obtain an equation with only one unknown (we must pay attention to the signs of subtraction and addition). If the coefficients are equal we will subtract the equations and if they are opposite we will add them.
  • Let's not forget to place the value found in one of the equations to discover the second unknown.

Practice Solving with the method of equalization for systems of two linear equations with two unknowns

Exercise #1

Solve the following equations:

(I)x+y=18 (I)x+y=18

(II)y=13 (II)y=13

Video Solution

Answer

x=5,y=13 x=5,y=13

Exercise #2

Solve the following equations:

(I)2x+y=9 (I)2x+y=9

(II)x=5 (II)x=5

Video Solution

Answer

x=5,y=1 x=5,y=-1

Exercise #3

Solve the following system of equations:

{xy=52x3y=8 \begin{cases} x-y=5 \\ 2x-3y=8 \end{cases}

Video Solution

Answer

x=2,y=3 x=2,y=-3

Exercise #4

Solve the above set of equations and choose the correct answer.

(I)2x+3y=4 (I)-2x+3y=4

(II)x4y=8 (II)x-4y=8

Video Solution

Answer

x=8,y=4 x=-8,y=-4

Exercise #5

Solve the above set of equations and choose the correct answer.

(I)5x+4y=3 (I)-5x+4y=3

(II)6x8y=10 (II)6x-8y=10

Video Solution

Answer

x=4,y=414 x=-4,y=-4\frac{1}{4}

Exercise #1

Find the value of x and and band the substitution method.

(I)x+and=5 (I)x+and=5

(II)2x3and=15 (II)2x-3and=-15

Video Solution

Answer

x=0,y=5 x=0,y=5

Exercise #2

Find the value of x and and band the substitution method.

(I)x2and=4 (I)-x-2and=4

(II)3x+and=8 (II)3x+and=8

Video Solution

Answer

x=4,y=4 x=4,y=-4

Exercise #3

Solve the above set of equations and choose the correct answer.

(I)8x+3y=7 (I)-8x+3y=7

(II)24x+y=3 (II)24x+y=3

Video Solution

Answer

x=0.025,y=2.4 x=0.025,y=2.4

Exercise #4

Solve the above set of equations and choose the correct answer.

(I)7x4y=8 (I)7x-4y=8

(II)x+5y=12.8 (II)x+5y=12.8

Video Solution

Answer

x=2.33,y=2.09 x=2.33,y=2.09

Exercise #5

Solve the following system of equations:

{8x+5y=310x+y=16 \begin{cases} -8x+5y=3 \\ 10x+y=16 \end{cases}

Video Solution

Answer

x=1.32,y=2.8 x=1.32,y=2.8

Exercise #1

Find the value of x and and band the substitution method.

(I)x+3and=12 (I)-x+3and=12

(II)4x+2and=10 (II)4x+2and=10

Video Solution

Answer

x=37,y=297 x=\frac{3}{7},y=\frac{29}{7}

Exercise #2

Find the value of x and and band the substitution method.

(I)5x+9and=18 (I)-5x+9and=18

(II)x+8and=16 (II)x+8and=16

Video Solution

Answer

x=0,y=2 x=0,y=2

Exercise #3

Solve the above set of equations and choose the correct answer.

(I)13x4y=5 (I)\frac{1}{3}x-4y=5

(II)x+6y=9 (II)x+6y=9

Video Solution

Answer

x=11,y=13 x=11,y=-\frac{1}{3}

Exercise #4

Solve the following system of equations:

{2x15y=183x+y=6 \begin{cases} 2x-\frac{1}{5}y=18 \\ 3x+y=6 \end{cases}

Video Solution

Answer

x=7.38,y=16.14 x=7.38,y=-16.14

Exercise #5

Find the value of x and and band the substitution method.

(I)4x+4and=15 (I)-4x+4and=15

(II)2x+8and=12 (II)2x+8and=12

Video Solution

Answer

x=95,y=3920 x=-\frac{9}{5},y=\frac{39}{20}

Topics learned in later sections

  1. Two linear equations with two unknowns
  2. Algebraic solution for linear equations with two unknowns
  3. Substitution method for two linear equations with two unknowns
  4. Linear equation with two variables