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(2x+5)(-mx+9)=0
In the given equation,
m is a constant. If the equation has the solutions
x=-(5)/(2) and
x=(3)/(2), what is the value of
m ?

$(2 x+5)(-m x+9)=0$ $\newline$ In the given equation, $m$ is a constant. If the equation has the solutions $x=-\frac{5}{2}$ and $x=\frac{3}{2}$ , what is the value of $m$ ?

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Write and solve direct variation equations

Full solution

Q.

(2 x+5)(-m x+9)=0

\newline

In the given equation,

m

is a constant. If the equation has the solutions

x=-\frac{5}{2}

and

x=\frac{3}{2}

, what is the value of

m

?

Given Equation and Solutions: We are given the equation $(2x+5)(-mx+9)=0$ and the solutions $x=-\frac{5}{2}$ and $x=\frac{3}{2}$ . Since these are solutions to the equation, we can substitute them into the equation to find the value of $m$ .
Substitute $x=-\frac{5}{2}$ : First, let's substitute $x=-\frac{5}{2}$ into the equation. $\newline$ $(2x+5)(-mx+9)=0$ $\newline$ $(2(-\frac{5}{2})+5)(-m(-\frac{5}{2})+9)=0$ $\newline$ $(-5+5)(\frac{5m}{2}+9)=0$ $\newline$ $0*(\frac{5m}{2}+9)=0$ $\newline$ Since multiplying by zero gives zero, this part of the equation is satisfied for any value of $m$ . This step does not help us find $m$ .
Substitute $x=\frac{3}{2}$ : Now, let's substitute $x=\frac{3}{2}$ into the equation. $\newline$ $(2x+5)(-mx+9)=0$ $\newline$ $\left(2\left(\frac{3}{2}\right)+5\right)\left(-m\left(\frac{3}{2}\right)+9\right)=0$ $\newline$ $\left(3+5\right)\left(-\frac{3m}{2}+9\right)=0$ $\newline$ $\left(8\right)\left(-\frac{3m}{2}+9\right)=0$ $\newline$ Since the first factor is $8$ and not zero, the second factor must be zero for the equation to hold true. $\newline$ $-\frac{3m}{2}+9=0$
Solve for m: Now we solve for m. $\newline$ $-\frac{3m}{2}+9=0$ $\newline$ $-\frac{3m}{2}=-9$ $\newline$ Multiply both sides by $-\frac{2}{3}$ to isolate m. $\newline$ $m=(-9)\cdot(-\frac{2}{3})$ $\newline$ $m=\frac{18}{3}$ $\newline$ $m=6$

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