The value of [tex]x[/tex] in the equation [tex]\frac{x}{5}-\frac{2y}{3}=30[/tex] for [tex]y=15[/tex] is [tex]\fbox{\begin\\\ \math x=200\\\end{minispace}}[/tex].
Further explanation:
The given equation is [tex]\frac{x}{5}-\frac{2y}{3}=30[/tex].
The given equation is a linear equation in two variables.
The general form of linear equation in two variables is [tex]ax+by+c=0[/tex] where [tex]a,b,c\ \text{and}\ c[/tex] are real numbers such that [tex]a\ \text{and}\ b[/tex] are not equal to zero.
The equation is [tex]\frac{x}{5}-\frac{2y}{3}=30[/tex].
Here, the value of [tex]y[/tex] is [tex]15[/tex].
The value of [tex]x[/tex] is calculated by substituting [tex]15[/tex] for [tex]y[/tex] as,
[tex]\dfrac{x}{5}-\dfrac{2\cdot 15}{3}=30[/tex]
Now, simplify the above equation as shown below.
[tex]\begin{aligned}\dfrac{x}{5}-\dfrac{2\cdot 15}{3}&=30\\ \dfrac{x}{5}-(2\cdot 5)&=30\\ \dfrac{x}{5}-10&=30\end{aligned}[/tex]
Now, add [tex]10[/tex] on each side of the above equation as,
[tex]\begin{aligned}\dfrac{x}{5}-10+10&=30+10\\ \dfrac{x}{5}&=40\end{aligned}[/tex]
The above expression can be further solved by multiplying [tex]5[/tex] on both side of the equation as,
[tex]\begin{aligned}\left(\dfrac{x}{5}\right)\cdot 5&=40\cdot 5\\x&=40\cdot 5\\x&=200\end{aligned}[/tex]
Therefore, the value of [tex]x[/tex] in the equation [tex]\frac{x}{5}-\frac{2y}{3}=30[/tex] for [tex]y=15[/tex] is [tex]\fbox{\begin\\\ \math x=200\\\end{minispace}}[/tex].
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Answer details
Grade: Middle school
Subject: Mathematics
Chapter: Linear equations in two variables
Keywords: Linear equations in one variable, linear equations in two variables, function, real numbers, solution, solution set, open interval, closed intervals, semi-closed intervals, semi-open interval, values, substitute, multiply, add, subtract, divide.
