Question

Solve the three algebraic expressions:
\[\begin{align}
  & 5x-6y+4z=15 \\
 & 7x+4y-3z=19 \\
 & 2x+y+6z=46 \\
\end{align}\]

Answer 1

Hint: Consider any 2 of the expressions and convert the 3 variables to 2 variables. Similarly find 2 equations and convert 3 variables to 2 variables.We get 2 equations having 2 variables and Solve them,we get values of two variables and substitute in any equation to get the value of another variable.Hence desired values of x ,y and z can be found.

Complete step-by-step answer:
We have been given three algebraic expressions. Now let us pick any two pairs of equations of the system.
\[\begin{align}
  & 5x-6y+4z=15-(1) \\
 & 7x+4y-3z=19-(2) \\
 & 2x+y+6z=46-(3) \\
\end{align}\]
Let us take equation (2) and equation (3). Then use addition or subtraction to eliminate the same variable from both pairs of equations.
\[\begin{align}
  & 7x+4y-3z=19-(2)\times 2 \\
 & 2x+y+6z=46-(3)\times 7 \\
\end{align}\]
Let us subtract both these equations and we will get an equation with 2 variables. Multiply 2 in equation (2) and 7 in equation (3).
14x + 8y - 6z = 38
14x + 7y + 42z = 322
            y – 48z = -284
\[\therefore y-48z=-284-(4)\]
Now let us take equation (1) and equation (3), multiply equation (3) by 5 and subtract both the equations.
\[\begin{align}
  & 5x-6y+4z=15-(1)\times 2 \\
 & 2x+y+6z=46-(3)\times 5 \\
\end{align}\]
Thus we get new equations as,
10x – 12y + 8z = 30
10x + 5y + 30z = 230
       - 17y -22z = -200
\[\Rightarrow 17y+22z=200-(5)\]
Now let us solve equation (4) and (5) to get the value of z.
\[\begin{align}
  & y-48z=-284-(4) \\
 & 17y+22z=200-(5) \\
\end{align}\]
Multiply equation (4) \[\times 17\].
Equation (4) becomes, \[17y-816z=-4828\].
17y – 816z = -4828
17y + 22z = 200
0 – 838z = -5028
\[\begin{align}
  & \therefore -838z=-5028 \\
 & \therefore z=\dfrac{5028}{838}=6 \\
\end{align}\]
Hence we got the value of z = 6, substitute this value in equation (4).
\[\begin{align}
  & y=48z-284=48\times 6-284=4 \\
 & \therefore y=4 \\
\end{align}\]
Substitute y = 4 and z = 6 in equation (1).
\[\begin{align}
  & 5x=15+6y-4z \\
 & 5x=15+\left( 6\times 4 \right)-\left( 4\times 6 \right) \\
 & \therefore x=\dfrac{15}{5}=3 \\
\end{align}\]
Thus we got the value as,
x = 3, y = 4 and z = 6.
Hence we solved the algebraic expressions.

Note: An algebraic expression can have any number of variables. Here we were given 3 variables x, y and z. Thus we required 3 equations to solve and find the values. For an expression of 2 variables only 2 equations are required.