In this problem, it is given that $4x+3y = 132$ and $5x-2y = -42$ are a system of linear equations and they should be solved mathematically by elimination method. So, let us learn how to solve the simultaneous linear system $4x+3y = 132$ and $5x-2y = -42$ in three simple steps by elimination.

In the linear equation $4x+3y = 132$, the coefficient of $y$ is $3$ but the coefficient of $y$ is $-2$ in the linear equation $5x-2y = -42$. It is not possible to eliminate the variable $y$ from them due to different numerical coefficients of $y$. So, we have to make them same for eliminating the terms that contain $y$. It is possible to achieve it mathematically.

Firstly, multiply $4x+3y = 132$ by $2$.

$2 \times (4x+3y) = 2 \times 132$

Use distributive property of multiplication over addition to multiply the factor $2$ by each term of the algebraic expression $4x+3y$ in the left hand side of the equation.

$\implies$ $2 \times 4x + 2 \times 3y = 2 \times 132$

$\,\,\, \therefore \,\,\,\,\,\,$ $8x+6y = 264$

Now multiply $5x-2y = -42$ by $3$.

$\implies$ $3 \times (5x-2y) = 3 \times (-42)$

Similarly, multiply the factor $3$ by each term of the algebraic expression $5x-2y$ as per distributive property of multiplication over subtraction.

$\implies$ $3 \times 5x -3 \times 2y = 3 \times (-42)$

$\,\,\, \therefore \,\,\,\,\,\,$ $15x -6y = -126$

Therefore, a system of linear equations $4x+3y = 132$ and $5x-2y = -42$ are transformed as $8x+6y = 264$ and $15x -6y = -126$ respectively.

The coefficient of $y$ is $6$ in linear equation in two variables $8x+6y = 264$ and the coefficient of $y$ is also $-6$ in linear equation $15x-6y = -126$. If the given system of linear equations in two variables are added, then the terms, which contains $y$, are eliminated and will be formed a linear equation in one variable. Then, it is easy to evaluate the variable $x$ from it.

$\implies$ $(8x+6y) + (15x-6y)$ $=$ $(264)+(-126)$

$\implies$ $8x+6y+15x-6y = 264-126$

$\implies$ $8x+15x+6y-6y = 138$

$\implies$ $\require{cancel} 23x+\cancel{6y}-\cancel{6y} = 138$

$\implies$ $23x = 222$

Now, use inverse operations method to solve the linear equation in one variable.

$\implies$ $\dfrac{23x}{23} = \dfrac{222}{23}$

$\implies$ $\require{cancel} \dfrac{\cancel{23}x}{\cancel{23}} = \dfrac{\cancel{222}}{\cancel{23}}$

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

Substitute the value of $x$ in any one of the given linear system in two variables $4x+3y = 132$ and $5x-2y = -42$. In this problem, the value of $x$ is substituted in $5x-2y = -42$.

$\implies$ $5(6)-2y = -42$ for calculating the value of $y$.

$\implies$ $5 \times 6-2y = -42$

$\implies$ $30-2y = -42$

$\implies$ $-2y = -30-42$

$\implies$ $-2y = -72$

$\implies$ $72 = 2y$

$\implies$ $2y = 72$

It is also a linear equation in one variable and solve it by transposing method.

$\implies$ $y = \dfrac{72}{2}$

$\implies$ $\require{cancel} y = \dfrac{\cancel{72}}{\cancel{2}}$

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

Therefore, it is solved that $x = 6$ and $y = 36$. In this way, a pair of simultaneous linear equations $4x+3y = 132$ and $5x-2y = -42$ are solved in mathematics by elimination method.

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