Question

What is the value of $a$ for the following equations $3a+4b=13\text{ and }a+3b=1$ ? (Use cross multiplication method).
A. $a=5$
B. $a=6$
C. $a=7$
D. $a=8$

Answer 1

Hint: To find the value of $a$ from $3a+4b=13\text{ and }a+3b=1$ , we will be using the formula
${\displaystyle \frac{x}{b_1{c}_2-b_2{c}_1}}={\displaystyle \frac{y}{c_1{a}_2-c_2{a}_1}}={\displaystyle \frac{1}{b_2{a}_1-b_1{a}_2}}$ for the equations $a_{1}x+b_{1}y+c_1=0\text{ and }{a}_{2}x+b_{2}y+c_2=0$ . We can compare this cross multiplication formula with the given two equations. When substituting these values, we will be getting an equation of the form ${\displaystyle \frac{a}{35}}={\displaystyle \frac{b}{-10}}={\displaystyle \frac{1}{5}}$ . From this, the value of $a$ can be obtained.

Complete step by step answer:
We need to find the value of $a$ from $3a+4b=13\text{ and }a+3b=1$ using the cross multiplication method.
Let us first recollect what cross multiplication method is.
Let us consider the equations $a_{1}x+b_{1}y+c_1=0\text{ and }{a}_{2}x+b_{2}y+c_2=0$
We can find the values of x and y are found using the cross multiplication formula shown below.
${\displaystyle \frac{x}{b_1{c}_2-b_2{c}_1}}={\displaystyle \frac{y}{c_1{a}_2-c_2{a}_1}}={\displaystyle \frac{1}{b_2{a}_1-b_1{a}_2}}...(i)$
Now, let us compare this with the given set of equations.
We have $3a+4b=13\text{ and }a+3b=1$
This can be written as $3a+4-13=0\text{ and }a+3b-1=0$
$\begin{array}{rl}& x=a,y=b\\ & a_1=3,a_2=1\\ & b_1=4,b_2=3\\ & c_1=-13,c_2=-1\end{array}$
Now, let us substitute the above values in equation (i). We will get
${\displaystyle \frac{a}{4\times -1-3\times (-13)}}={\displaystyle \frac{b}{-13\times 1-(-1)\times 3}}={\displaystyle \frac{1}{3\times 3-4\times 1}}$
Let us simplify the denominators. We will get
${\displaystyle \frac{a}{-4+39}}={\displaystyle \frac{b}{-13+3}}={\displaystyle \frac{1}{9-4}}$
$\Rightarrow {\displaystyle \frac{a}{35}}={\displaystyle \frac{b}{-10}}={\displaystyle \frac{1}{5}}$
Now let us consider the first and last terms, that is,
${\displaystyle \frac{a}{35}}={\displaystyle \frac{1}{5}}$
This can be written as
$a={\displaystyle \frac{1\times 35}{5}}$
When we solve this, we get
$a=7$

So, the correct answer is “Option C”.

Note: Be careful when substituting the equations in the standard formula. The given equations must be converted to the standard form $a_{1}x+b_{1}y+c_1=0\text{ and }{a}_{2}x+b_{2}y+c_2=0$ . You must be cautious with the denominators when writing the formula ${\displaystyle \frac{x}{b_1{c}_2-b_2{c}_1}}={\displaystyle \frac{y}{c_1{a}_2-c_2{a}_1}}={\displaystyle \frac{1}{b_2{a}_1-b_1{a}_2}}$ .
We can see that the denominator of x does not contain its coefficient. Also the subscripts of the terms of the denominator are different. For example, $b_1{c}_2\text{ and }{b}_2{c}_1$ .