$4x+3y = 132$ and $5x-2y = -42$ are a given system of linear equations in two variables. In this problem, it is given that the solution of the both simultaneous linear equations in two variables have to calculate in the substitution method. So, let us learn how to solve the pair of the given linear equations in two variables in three simple steps by the substitution.

The values of both variables $x$ and $y$ are unknown in the given system of linear equations. In substitution method, the value of one variable is expressed in terms of the remaining variable by considering any one of the linear equations. In this problem, $4x+3y = 132$ is considered and express the value of $x$ in terms of $y$.

$\implies$ $4x = -3y+132$

$\implies$ $x = \dfrac{-3y+132}{4}$

$\implies$ $x = \dfrac{-3y}{4}+\dfrac{132}{4}$

$\implies$ $x = \require{cancel} -\dfrac{3y}{4}+\dfrac{\cancel{132}}{\cancel{4}}$

$\,\,\, \therefore \,\,\,\,\,\,$ $x = -\dfrac{3y}{4}+33$

In the above step, the value of $x$ is written as $-\dfrac{3y}{4}+33$ and substitute it in the second linear equation $5x-2y = -42$ for converting this linear equation in two variables into linear equation in one variable.

$\implies$ $5\Bigg(-\dfrac{3y}{4}+33\Bigg)-2y = -42$

Now, use distributive property of multiplication across subtraction to distribute the factor $5$ to the difference of the terms.

$\implies$ $5 \times \Bigg(-\dfrac{3y}{4}\Bigg)+5 \times 33-2y = -42$

Now, simplify the equation in algebraic form.

$\implies$ $-\dfrac{5 \times 3y}{4}+165-2y = -42$

$\implies$ $-\dfrac{15y}{4}-2y = -165-42$

$\implies$ $-\dfrac{15y}{4}-2y = -207$

$\implies$ $207 = \dfrac{15y}{4}+2y$

$\implies$ $\dfrac{15y}{4}+2y = 207$

Multiply both sides of the equation by number $4$ to eliminate the $4$ from the denominator in the first term of the left hand side of the equation.

$\implies$ $4 \times \Bigg(\dfrac{15y}{4}+2y\Bigg) = 4 \times 207$

Multiply the factor $4$ by the sum of the terms as per the distributive property of multiplication over addition.

$\implies$ $4 \times \dfrac{15y}{4}+4 \times 2y = 828$

$\implies$ $\dfrac{4 \times 15y}{4}+8y = 828$

$\implies$ $\require{cancel} \dfrac{\cancel{4} \times 15y}{\cancel{4}}+8y = 828$

$\implies$ $15y+8y = 828$

$\implies$ $23y = 828$

It is a linear equation in one variable and solve it to evaluate $y$ by transposing.

$\implies$ $y = \dfrac{828}{23}$

$\implies$ $\require{cancel} y = \dfrac{\cancel{828}}{\cancel{23}}$

$\,\,\, \therefore \,\,\,\,\,\,$ $y = 36$

Therefore, it is solved that the value of variable $y$ is equal to $36$.

For calculating the variable $x$, substitute the value of $y$ in any one of the equation of the linear system. In this problem, substitute $y = 36$ in the linear equation $4x+3y = 132$.

$\implies$ $4x+3(36) = 132$

$\implies$ $4x+3 \times 36 = 132$

$\implies$ $4x+108 = 132$

$\implies$ $4x = 132-108$

$\implies$ $4x = 24$

$\implies$ $x = \dfrac{24}{4}$

$\implies$ $\require{cancel} x = \dfrac{\cancel{24}}{\cancel{4}}$

$\,\,\, \therefore \,\,\,\,\,\,$ $x = 6$

Therefore, it is solved that the value of variable $x$ is equal to $6$.

Thus, a pair of simultaneous linear equations in two variables $4x+3y = 132$ and $5x-2y = -42$ are solved mathematically by substitution method.

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