Algebraic solution - Examples, Exercises and Solutions

A system of linear equations is, in fact, a set of conditions that must be satisfied by specific unknowns,
the solution for the system of equations is then based on finding the X X and the Y Y that agree with both the first equation and the second equation.

These questions can be solved in several ways, the algebraicsolution includes two methods:

Substitution method:

  1. Isolate an unknown in any of the equations.
  2. Substitute the unknown we isolated in the other equation of the system and find the value of an unknown.
  3. We place the value of the unknown that we have discovered in one equation to find the value of the other.

Equalization method

  1. We will cause the coefficients in both equations (X X or Y Y ) to become equal.
  2. We will add or subtract one equation from the other and thus eliminate the equal coefficients.
  3. We will solve the equation with the isolated coefficient and find its value.
  4. We place the value of the unknown that we have discovered in one equation to find the value of the other.

Practice Algebraic solution

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. Substitution method for two linear equations with two unknowns
  3. Solving with the method of equalization for systems of two linear equations with two unknowns
  4. Linear equation with two variables