How to Solve Algebraic Expressions Step by Step

Solving algebraic expression means finding the value of the variable. To solve the linear equations we need to do some algebraic operation to find out the value of one of the variables and then by substituting that value we can get the value of another variable.

How to Solve One Variable Linear Equations

To solve the one variable linear equation we do use the transposition method. According to this method, perform the same operation on both sides of equality. In short, perform the operations to isolate variables.

Examples:

1. 5x + 7 = 32

Subtracting 7 from both sides we have

⇒ 5x = 25

Dividing both sides by 5 we have

⇒ x = 5

2. 3(2y – 12) = 72

Dividing both sides by 3 we have

⇒ 2y – 12 = 24

Adding 12 in both sides we have

⇒ 2y = 36

Dividing both sides by 2 we have

⇒ y = 18.

How to Solve Two Variable Linear Equations

To solve the two variables linear equation we have some methods given below:

Elimination Method Steps with Example

Let’s suppose we are having two equations

a1 x + b1 y + c1 = 0 ….(1)

a2 x + b2 y + c2 = 0 …..(2)

In this method, make the coefficient of any one of the variables equal.

Multiplying equation (1) with a2 and equation (2) with a1 , we have

⇒ a1 a2 x + b1 a2 y + a2 c1 = 0 ….(3)

⇒ a2 a1 x +a1 b2 y + a1 c2 = 0 ….(4)

Subtracting equation (4) from equation (3) we have

⇒ (b1 a2 – a1 b2)y + a2 c– a_1 c=0

⇒ y = $\frac{a_1 c_2 – a_2 c_1}{b_1 a_2-a_1 b_2 }$.

Now by substituting the value of y in any one of the equations we can get the value of x.

Example:

For the given equations find values of x and y

2x + 3y – 6 = 0 ..(1)

3x – 4y + 2 = 0 ..(2)

Multiplying equation (1) and (2) with 3 and 2 respectively we have

⇒ 6x + 9y – 18 = 0 ….(3)

⇒ 6x – 8y + 4 = 0 ….(4)

Subtracting equation (4) from equation (3) we have

⇒ 17y – 22 = 0

⇒ y = 22/17.

Substituting this value in equation (1) we have

⇒ 2x + 3(22/17) – 6 = 0

⇒ 2x +66/17- 6 = 0

⇒ x +44/17- 3 = 0

⇒ x = (51 – 44)/17

⇒ x = 7/17.

Substitution method Step by Step with Example:

Let’s suppose we are having two equations

a1 x + b1 y + c1 = 0 ….(1)

a2 x + b2 y + c2 = 0 …..(2)

From equation (2) we can find the value of y in terms of x and by substituting that value in equation (1) we can find the value of x.

Example:

For the given equations find values of x and y

5x – 7y + 21 = 0 ..(1)

x + y – 3 = 0 ..(2)

From equation (2) we have

⇒ x = 3 – y

Substituting this value in equation (1) we have

⇒ 5(3 – y) – 7y + 21 = 0

⇒ 15 – 5y -7y + 21 = 0

⇒ 12y = 36

⇒ y = 3.

⇒ x = 3 – y

⇒ x = 3 – 3

⇒ x = 0.

Cross multiplication method Step by Step with Example:

Let’s suppose we are having two equations

a1 x + b1 y + c1 = 0 ….(1)

a2 x + b2 y + c2 = 0 …..(2)

According to cross multiplication method we can use the result given below:

$\frac{x}{b_1 c_2-c_1 b_2 }\ =\ \frac{y}{c_1 a_2-a_1 c_2 }\ =\ \frac{1}{a_1 b_2-b_1 a_2 }$.

By using this result we can find the value of x and y.

Example:

For the given equations find values of x and y

2x – y + 3 = 0 ..(1)

x – y – 2 = 0 ..(2)

Using the result $\frac{x}{b_1 c_2-c_1 b_2 }\ =\ \frac{y}{c_1 a_2-a_1 c_2 }\ =\ \frac{1}{a_1 b_2-b_1 a_2 }$, we have

$\frac{x}{2 – (-3)}\ =\ \frac{y}{3 – (-4)}\ =\ \frac{1}{-2 – (-1)}$

⇒ x/(2 + 3) = y/(3 + 4) = 1/(-2 + 1)

⇒ x/5 = y/7 = 1/(-1)

⇒ x/5 = 1/(-1) and y/7 = 1/(-1)

⇒ x = – 5 and y = – 7.

Roots of Complex Algebraic Equations

To solve the complex algebraic equation we have to equate the real parts and imaginary parts.

If a + ib = c + id, then a = c and b = d.

By using this result we can solve linear equations of 2 variables.

Example:

For the given equation find values of x and y

x – 2y + 3 + i(-2x – y – 2) = – 3x – y – i(3x)

Equating real and imaginary parts we have

⇒ x – 2y + 3 = – 3x – y and -2x – y – 2 = – 3x

⇒ 2x – y + 3 = 0 and x – y – 2 = 0

⇒ 2x – y + 3 = 0 (1)

⇒ x – y – 2 = 0 (2)

Using the result $\frac{x}{2 – (-3)}\ =\ \frac{y}{3 – (-4)}\ =\ \frac{1}{-2 – (-1)}$, we have

⇒ x/(2 – (-3))=y/(3 – (-4))=1/(-2 – (-1))

⇒ x/(2 + 3)=y/(3 + 4)=1/(-2 + 1)

⇒ x/5=y/7=1/(-1)

⇒ x/5= 1/(-1) and y/7=1/(-1)

⇒ x = – 5 and y = – 7.

How to Solve 3 Variables Simultaneous Equations

To solve 3 variables simultaneous linear equations we can use elimination and substitution methods.

Let’s suppose we are having three equations

a1 x + b1 y + c1 z = 0 ..(1)

a2 x + b2 y + c2 z = 0 …(2)

a3 x + b3 y + c3 z = 0 …(3)

Finding the value of variable z in terms of x and y from equation (3) and substituting it in the equation (1) and (2), we can convert first two equations in two variables and then solving using elimination method we can find the value of x and y and hence we can find value of z.

How to Solve Three Variable Equations by Cross Multiplication

To solve three variable equations by cross multiplication we have to find the value of first two variables in terms of third variable by cross multiplication method in two equations and then by substituting those value in third equation we can get the value of third variable and hence value of other variables.

Note: We can apply this method only when at least any two of three equations are homogeneous i.e. constant part is zero and we have to apply cross multiplication in those two equations only.

Let’s suppose we are having three equations

a1 x + b1 y + c1 z = 0 ..(1)

a2 x + b2 y + c2 z = 0 …(2)

a3 x + b3 y + c3 z = 0 …(3)

According to cross multiplication method we can use the result given below:

$\frac{x}{b_1 c_2-c_1 b_2 }= \frac{y}{c_1 a_2-a_1 c_2 } = \frac{z}{a_1 b_2-b_1 a_2 }$.

Using this result we can find the value of x and y in terms of z and then by substituting that value in equation (3) we can find value of z and hence values of x and y.

Algebraic expressions and equations problems

Solved Problems

Example 1:

Find the values of x,y and z for given equations

2x – y + 3 = 0 ..(1)

x – y – 2 = 0 …(2)

x + 2y + z + 36 = 0 …(3)

Solution: Using the result $\frac{x}{b_1 c_2-c_1 b_2 }= \frac{y}{c_1 a_2-a_1 c_2 } = \frac{z}{a_1 b_2-b_1 a_2 }$ in first two equations we have

⇒ x/(2 – (-3))=y/(3 – (-4))=z/(-2 – (-1))

⇒ x/(2 + 3)=y/(3 + 4)=z/(-2 + 1)

⇒ x/5=y/7=z/(-1)

⇒ x/5= z/(-1) and y/7=z/(-1)

⇒ x = – 5z and y = – 7z

Substituting these values in equation (3) we have

⇒ -5z – 14z + z +36 = 0

⇒ z = 2

We know x = – 5z and y = – 7z

⇒ x = -10 and y = -14.

Example 2:

Find the values of x,y and z for given equations

2x + 3y – 5z + 5 = 0 …… (1)

x – y – z +1 = 0 ……. (2)

4x – 2y + z + 2 = 0 ………(3)

Solution:

From equation (3) we have

⇒ z = -4x + 2y – 2

Substituting this value in equations (1) and (2) we have

⇒ 2x + 3y – 5(-4x + 2y – 2) + 5 = 0 and x – y – (-4x + 2y – 2) +1 = 0

⇒ 2x + 3y + 20x -10y + 10 + 5 = 0 and x – y + 4x – 2y + 2 + 1 = 0

⇒ 22x – 7y + 15 = 0 and 5x – 3y + 3 = 0

⇒ 5x – 3y + 3 = 0 (4)

⇒ 22x – 7y + 15 = 0 (5)

Multiplying equations (4) and (5) by 7 and 3 respectively we have

⇒ 35x – 21y + 21 = 0 (6)

⇒ 66x – 21y + 45 = 0 (7)

Subtracting equation (6) from equation (7) we have

⇒ 31x + 24 = 0

⇒ x = 24/31.

Substituting this value in equation (4) we have

⇒ 5(24/31) – 3y + 3 = 0

⇒ 3y = 120/31+ 3

⇒ 3y = (120 + 93)/31

⇒ 3y = 213/31

⇒ y = 71/31.

We know

⇒ z = -4x + 2y – 2

Substituting values of x and y we have

⇒ z = -4(24/31) + 2(71/31) – 2

⇒ z = -96/31 + 142/31 – 2

⇒ z = (-96 + 142 – 62)/31

⇒ z = -16/31.

Example 3: Find the values of x,y, and z for given equations

x + y + z = 25,

5x + 3y + 2z = 0,

y – z = 6

Solution:

$\begin{bmatrix}x+y+z=25\\ 5x+3y+2z=0\\ y-z=6\end{bmatrix}\\ \mathrm{Multiply\:}x+y+z=25\mathrm{\:by\:}5:\quad 5x+5y+5z=125\\ {[5x+3y+2z=0]} – {[5x+5y+5z=125]} = {[-2y-3z=-125]}\\ \begin{bmatrix}5x+5y+5z=125\\ -2y-3z=-125\\ y-z=6\end{bmatrix}\\ \mathrm{Multiply\:}y-z=6\mathrm{\:by\:}2:\quad 2y-2z=12\\ \begin{bmatrix}5x+5y+5z=125\\ -2y-3z=-125\\ 2y-2z=12\end{bmatrix}\\ {[2y-2z=12]} – {[-2y-3z=-125]} = {[-5z=-113]}\\ \begin{bmatrix}5x+5y+5z=125\\ -2y-3z=-125\\ -5z=-113\end{bmatrix}\\ \mathrm{Solve}\:-5z=-113\:\mathrm{for}\:z:\quad z=\frac{113}{5}\\ \mathrm{Solve}\:-2y-3\frac{113}{5}=-125\:\mathrm{for}\:y:\quad y=\frac{143}{5}\\ \mathrm{Solve}\:5x+5\frac{143}{5}+5\frac{113}{5}=125\:\mathrm{for}\:x:\quad x=-\frac{131}{5}\\$

The solution to the system of equations is:

$z=\frac{113}{5},\:y=\frac{143}{5},\:x=-\frac{131}{5}$

Example 4: Solve the given equation for x, y and z values.

2x – y + 3z = 9

x + y + z = 6

x – y + z = 2

Solution:

$\begin{bmatrix}2x-y+3z=9\\ x+y+z=6\\ x-y+z=2\end{bmatrix}\\\mathrm{Isolate}\:x\:\mathrm{for}\:2x-y+3z=9:\quad x=\frac{9+y-3z}{2}\\\mathrm{Subtract\:}-y+3z\mathrm{\:from\:both\:sides}\\2x-y+3z-\left(-y+3z\right)=9-\left(-y+3z\right)\\\mathrm{Simplify}\\2x=9+y-3z\\\mathrm{Divide\:both\:sides\:by\:}2\\\frac{2x}{2}=\frac{9}{2}+\frac{y}{2}-\frac{3z}{2}\\\mathrm{Simplify}\\x=\frac{9+y-3z}{2}\\\mathrm{Substitute\:}x=\frac{9+y-3z}{2}\\\begin{bmatrix}\frac{9+y-3z}{2}+y+z=6\\ \frac{9+y-3z}{2}-y+z=2\end{bmatrix}\\\mathrm{Isolate}\:y\:\mathrm{for}\:\frac{9+y-3z}{2}+y+z=6:\quad y=\frac{z+3}{3}\\\mathrm{Substitute\:}y=\frac{z+3}{3}\\\begin{bmatrix}\frac{9+\frac{z+3}{3}-3z}{2}-\frac{z+3}{3}+z=2\end{bmatrix}\\\mathrm{Isolate}\:z\:\mathrm{for}\:\frac{9+\frac{z+3}{3}-3z}{2}-\frac{z+3}{3}+z=2:\quad z=3\\\mathrm{For\:}y=\frac{z+3}{3}\\\mathrm{Substitute\:}z=3\\y=\frac{3+3}{3}\\y=2\\\mathrm{For\:}x=\frac{9+y-3z}{2}\\\mathrm{Substitute\:}z=3\\\mathrm{Substitute\:}y=2\\x=\frac{9+2-3\cdot \:3}{2}\\x=1\\\mathrm{The\:solutions\:to\:the\:system\:of\:equations\:are:}\\z=3,\:y=2,\:x=1\\$

Example 5: Solve the following set of equations for x,y and z.

5x + 3y – 2z = 7

3x – 5y + 6z = -23

15x + 2y – 4z = 9

Solution:

$\begin{bmatrix}5x+3y-2z=7\\ 3x-5y+6z=-23\\ 15x+2y-4z=9\end{bmatrix}\\\mathrm{Isolate}\:x\:\mathrm{for}\:5x+3y-2z=7:\quad x=\frac{7-3y+2z}{5}\\\mathrm{Substitute\:}x=\frac{7-3y+2z}{5}\\\begin{bmatrix}3\cdot \frac{7-3y+2z}{5}-5y+6z=-23\\ 15\cdot \frac{7-3y+2z}{5}+2y-4z=9\end{bmatrix}\\\mathrm{Isolate}\:y\:\mathrm{for}\:3\frac{7-3y+2z}{5}-5y+6z=-23:\quad y=\frac{2\left(9z+34\right)}{17}\\\mathrm{Substitute\:}y=\frac{2\left(9z+34\right)}{17}\\\begin{bmatrix}15\cdot \frac{7-3\cdot \frac{2\left(9z+34\right)}{17}+2z}{5}+2\cdot \frac{2\left(9z+34\right)}{17}-4z=9\end{bmatrix}\\\mathrm{Isolate}\:z\:\mathrm{for}\:15\frac{7-3\frac{2\left(9z+34\right)}{17}+2z}{5}+2\frac{2\left(9z+34\right)}{17}-4z=9:\quad z=-\frac{68}{23}\\\mathrm{For\:}y=\frac{2\left(9z+34\right)}{17}\\\mathrm{Substitute\:}z=-\frac{68}{23}\\y=\frac{2\left(9\left(-\frac{68}{23}\right)+34\right)}{17}\\y=\frac{20}{23}\\\mathrm{For\:}x=\frac{7-3y+2z}{5}\\\mathrm{Substitute\:}z=-\frac{68}{23}\\\mathrm{Substitute\:}y=\frac{20}{23}\\x=\frac{7-3\cdot \frac{20}{23}+2\left(-\frac{68}{23}\right)}{5}\\x=-\frac{7}{23}\\\mathrm{The\:solutions\:to\:the\:system\:of\:equations\:are:}\\z=-\frac{68}{23},\:y=\frac{20}{23},\:x=-\frac{7}{23}\\$