Question

How do you solve the system of equations $8x + 8y = - 8$ and $8x + 3y = 17$ using elimination?

Answer 1

Hint: First, multiply the equations so as to make the coefficients of the variables to be eliminated equal. Next, add or subtract the equation the equations obtained according as the terms having the same coefficients are of opposite or of the same sign. Solve the equation in one variable. Substitute the value found in any of the equations after substitution and find the value of another variable. The values of the variables constitute the solution of the given system of equations.
Formula used:
Method of Elimination:
In this method, we eliminate one of the two variables to obtain an equation in one variable which can be easily solved. Putting the value of this variable in any one of the given equations, the value of another variable can be obtained.

Complete step-by-step solution:
The given system of equations is
$8x + 8y = - 8$…(i)
$8x + 3y = 17$…(ii)
Now, multiply the equations so as to make the coefficients of the variables to be eliminated equal.
Here, the coefficient of the variable $x$ is already equal.
Next, add or subtract the equation the equations obtained according as the terms having the same coefficients are of opposite or of the same sign.
We can eliminate variable $x$ by subtracting equations (ii) from (i).
$5y = - 25$
Divide both sides of equation by $5$, we get
$y = - 5$
Now, substitute the value of $y$ in equation (i) and find the value of $x$.
$8x + 8\left( { - 5} \right) = - 8$
$ \Rightarrow 8x - 40 = - 8$
Add $40$ to both sides of the equation, we get
$ \Rightarrow 8x = 32$
Divide both sides of equation by $8$, we get
$\therefore x = 4$
Hence, the solution of the given system of equations is $x = 4$, $y = - 5$.

Note: We can also find the solution of a given system by Method of Cross-Multiplication.
System of equations:
$8x + 8y = - 8$…(i)
$8x + 3y = 17$…(ii)
By cross-multiplication, we have
$\dfrac{x}{{\begin{array}{*{20}{c}}
  8&8 \\
  3&{ - 17}
\end{array}}} = \dfrac{{ - y}}{{\begin{array}{*{20}{c}}
  8&8 \\
  8&{ - 17}
\end{array}}} = \dfrac{1}{{\begin{array}{*{20}{c}}
  8&8 \\
  8&3
\end{array}}}$
$ \Rightarrow \dfrac{x}{{ - 136 - 24}} = \dfrac{{ - y}}{{ - 136 - 64}} = \dfrac{1}{{24 - 64}}$
$ \Rightarrow \dfrac{x}{{ - 160}} = \dfrac{{ - y}}{{ - 200}} = \dfrac{1}{{ - 40}}$
$ \Rightarrow x = \dfrac{{ - 160}}{{ - 40}} = 4$ and $y = \dfrac{{200}}{{ - 40}} = - 5$
Final solution: Hence, the solution of the given system of equations is $x = 4$, $y = - 5$.