$12t = 4v - 3$ and $-6t = 4v + 6$ . If $(t, v)$ is the solution to the system of equations, what is the value of $t - v$ ?

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Grade 6

Evaluate variable expressions with decimals, fractions, and mixed numbers

Full solution

12t = 4v - 3

and

-6t = 4v + 6

. If

(t, v)

is the solution to the system of equations, what is the value of

t - v

Write Equations: Write down the system of equations. $\newline$ We have the following system of equations: $\newline$ $12t = 4v - 3$ $\newline$ $-6t = 4v + 6$
Modify Second Equation: Multiply the second equation by $2$ to make the coefficients of $t$ the same in both equations. $\newline$ Multiplying the second equation by $2$ gives us: $\newline$ $-12t = 8v + 12$
Eliminate Variable: Add the modified second equation to the first equation to eliminate $t$ . Adding the equations we get: $12t + (-12t) = (4v - 3) + (8v + 12)$ $0 = 12v + 9$
Solve for $v$ : Solve for $v$ . $\newline$ Subtract $9$ from both sides to isolate the term with $v$ : $\newline$ $12v = -9$ $\newline$ Divide both sides by $12$ to solve for $v$ : $\newline$ $v = -\frac{9}{12}$ $\newline$ Simplify the fraction: $\newline$ $v = -\frac{3}{4}$
Substitute and Solve for t: Substitute the value of $v$ into one of the original equations to solve for $t$ . Let's use the first equation: $12t = 4v - 3$ Substitute $v$ with $-\frac{3}{4}$ : $12t = 4(-\frac{3}{4}) - 3$
Calculate $t - v$ : Simplify the equation and solve for $t$ .
$12t = -3 - 3$
$12t = -6$
Divide both sides by $12$ to solve for $t$ :
$t = \frac{-6}{12}$
Simplify the fraction:
$t = \frac{-1}{2}$
Calculate $t - v$ : Simplify the equation and solve for $t$ . $\newline$ $12t = -3 - 3$ $\newline$ $12t = -6$ $\newline$ Divide both sides by $12$ to solve for $t$ : $\newline$ $t = -\frac{6}{12}$ $\newline$ Simplify the fraction: $\newline$ $t = -\frac{1}{2}$ Calculate $t - v$ . $\newline$ $t - v = (-\frac{1}{2}) - (-\frac{3}{4})$ $\newline$ To subtract fractions, find a common denominator. In this case, the common denominator is $t$ $0$ : $\newline$ $t$ $1$ $\newline$ $t$ $2$ $\newline$ $t$ $3$