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$Find U_(1), U_(2), U_(3), U_(4). {:[U_(1)+U_(2)+U_(3)=220 v],[U_(1)+U_(4)=200 v],[U_(2)+U_(3)=140 v],[U_(3)+U_(4)=160 v]:}$

Find $U_{1}, U_{2}, U_{3}, U_{4}$ . $\newline$ $\begin{array}{l}U_{1}+U_{2}+U_{3}=220 v \\ U_{1}+U_{4}=200 v \\ U_{2}+U_{3}=140 v \\ U_{3}+U_{4}=160 v\end{array}$

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Math Problems

Grade 7

Solve two-step equations with parentheses

Full solution

Q. Find

U_{1}, U_{2}, U_{3}, U_{4}

\newline

\begin{array}{l}U_{1}+U_{2}+U_{3}=220 v \\ U_{1}+U_{4}=200 v \\ U_{2}+U_{3}=140 v \\ U_{3}+U_{4}=160 v\end{array}

Write Equations: Write down the system of equations. $\newline$ We have the following system of equations: $\newline$ $1$ ) $U_1 + U_2 + U_3 = 220 \, \text{V}$ $\newline$ $2$ ) $U_1 + U_4 = 200 \, \text{V}$ $\newline$ $3$ ) $U_2 + U_3 = 140 \, \text{V}$ $\newline$ $4$ ) $U_3 + U_4 = 160 \, \text{V}$
Subtract to Find U₁: Subtract equation $3$ from equation $1$ to find U₁. $\newline$ By subtracting $U_2 + U_3$ from both sides of equation $1$ , we get: $\newline$ $U_1 + U_2 + U_3 - (U_2 + U_3) = 220 \, \text{V} - 140 \, \text{V}$ $\newline$ $U_1 = 80 \, \text{V}$
Find $U_4$ : Substitute $U_1$ into equation $2$ to find $U_4$ . We know $U_1 = 80\,\text{V}$ , so we substitute it into equation $2$ : $U_1 + U_4 = 200\,\text{V}$ $80\,\text{V} + U_4 = 200\,\text{V}$ $U_4 = 200\,\text{V} - 80\,\text{V}$ $U_4 = 120\,\text{V}$
Find $U_2$ and $U_3$ : Substitute $U_1$ into equation $1$ and then use equation $3$ to find $U_2$ and $U_3$ . We know $U_1 = 80\,\text{V}$ , so we substitute it into equation $1$ and use equation $3$ : $U_1 + U_2 + U_3 = 220\,\text{V}$ $80\,\text{V} + U_2 + U_3 = 220\,\text{V}$ $U_2 + U_3 = 220\,\text{V} - 80\,\text{V}$ $U_2 + U_3 = 140\,\text{V}$ Since $U_2 + U_3 = 140\,\text{V}$ is the same as equation $3$ , we do not have new information here. We need to use another equation to find $U_2$ and $U_3$ .
Find $U_3$ : Substitute $U_4$ into equation $4$ to find $U_3$ . We know $U_4 = 120$ V, so we substitute it into equation $4$ : $U_3 + U_4 = 160$ V $U_3 + 120$ V = $160$ V $U_3 = 160$ V - $120$ V $U_3 = 40$ V
Find $U_2$ : Substitute $U_3$ into equation $3$ to find $U_2$ . We know $U_3 = 40\,\text{V}$ , so we substitute it into equation $3$ : $U_2 + U_3 = 140\,\text{V}$ $U_2 + 40\,\text{V} = 140\,\text{V}$ $U_2 = 140\,\text{V} - 40\,\text{V}$ $U_2 = 100\,\text{V}$