Which system of equations has a solution of approximately $$ (-2.7,-1.2) $$ ? $$ 4 x-5 y=-5 $$ and $$ 3 x+10 y=-20 $$ $$ 4 x-5 y=5 $$ and $$ 3 x+10 y=20 $$ $$ 4 x-5 y=5 $$ and $$ 3 x+10 y=-20 $$ $$ 4 x-5 y=-5 $$ and $$ 3 x+10 y=20 $$ Mark this and return

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Solve the system of equations by following steps:
- step0: Solve using the substitution method:
  $$\left\{ \begin{array}{l}4x-5y=-5\3x+10y=-20\end{array}\right.$$
- step1: Solve the equation:
  $$\left\{ \begin{array}{l}x=\frac{-5+5y}{4}\3x+10y=-20\end{array}\right.$$
- step2: Substitute the value of $$x:$$
  $$3\times \frac{-5+5y}{4}+10y=-20$$
- step3: Simplify:
  $$\frac{3\left(-5+5y\right)}{4}+10y=-20$$
- step4: Multiply both sides of the equation by LCD:
  $$\left(\frac{3\left(-5+5y\right)}{4}+10y\right)\times 4=-20\times 4$$
- step5: Simplify the equation:
  $$-15+55y=-80$$
- step6: Move the constant to the right side:
  $$55y=-80+15$$
- step7: Add the numbers:
  $$55y=-65$$
- step8: Divide both sides:
  $$\frac{55y}{55}=\frac{-65}{55}$$
- step9: Divide the numbers:
  $$y=-\frac{13}{11}$$
- step10: Substitute the value of $$y:$$
  $$x=\frac{-5+5\left(-\frac{13}{11}\right)}{4}$$
- step11: Calculate:
  $$x=-\frac{30}{11}$$
- step12: Calculate:
  $$\left\{ \begin{array}{l}x=-\frac{30}{11}\y=-\frac{13}{11}\end{array}\right.$$
- step13: Check the solution:
  $$\left\{ \begin{array}{l}x=-\frac{30}{11}\y=-\frac{13}{11}\end{array}\right.$$
- step14: Rewrite:
  $$\left(x,y\right) = \left(-\frac{30}{11},-\frac{13}{11}\right)$$
Solve the system of equations $$ 4 x - 5 y = -5; 3 x + 10 y = 20 $$.
Solve the system of equations by following steps:
- step0: Solve using the substitution method:
  $$\left\{ \begin{array}{l}4x-5y=-5\3x+10y=20\end{array}\right.$$
- step1: Solve the equation:
  $$\left\{ \begin{array}{l}x=\frac{-5+5y}{4}\3x+10y=20\end{array}\right.$$
- step2: Substitute the value of $$x:$$
  $$3\times \frac{-5+5y}{4}+10y=20$$
- step3: Simplify:
  $$\frac{3\left(-5+5y\right)}{4}+10y=20$$
- step4: Multiply both sides of the equation by LCD:
  $$\left(\frac{3\left(-5+5y\right)}{4}+10y\right)\times 4=20\times 4$$
- step5: Simplify the equation:
  $$-15+55y=80$$
- step6: Move the constant to the right side:
  $$55y=80+15$$
- step7: Add the numbers:
  $$55y=95$$
- step8: Divide both sides:
  $$\frac{55y}{55}=\frac{95}{55}$$
- step9: Divide the numbers:
  $$y=\frac{19}{11}$$
- step10: Substitute the value of $$y:$$
  $$x=\frac{-5+5\times \frac{19}{11}}{4}$$
- step11: Calculate:
  $$x=\frac{10}{11}$$
- step12: Calculate:
  $$\left\{ \begin{array}{l}x=\frac{10}{11}\y=\frac{19}{11}\end{array}\right.$$
- step13: Check the solution:
  $$\left\{ \begin{array}{l}x=\frac{10}{11}\y=\frac{19}{11}\end{array}\right.$$
- step14: Rewrite:
  $$\left(x,y\right) = \left(\frac{10}{11},\frac{19}{11}\right)$$
Solve the system of equations $$ 4 x - 5 y = 5; 3 x + 10 y = 20 $$.
Solve the system of equations by following steps:
- step0: Solve using the substitution method:
  $$\left\{ \begin{array}{l}4x-5y=5\3x+10y=20\end{array}\right.$$
- step1: Solve the equation:
  $$\left\{ \begin{array}{l}x=\frac{5+5y}{4}\3x+10y=20\end{array}\right.$$
- step2: Substitute the value of $$x:$$
  $$3\times \frac{5+5y}{4}+10y=20$$
- step3: Simplify:
  $$\frac{3\left(5+5y\right)}{4}+10y=20$$
- step4: Multiply both sides of the equation by LCD:
  $$\left(\frac{3\left(5+5y\right)}{4}+10y\right)\times 4=20\times 4$$
- step5: Simplify the equation:
  $$15+55y=80$$
- step6: Move the constant to the right side:
  $$55y=80-15$$
- step7: Subtract the numbers:
  $$55y=65$$
- step8: Divide both sides:
  $$\frac{55y}{55}=\frac{65}{55}$$
- step9: Divide the numbers:
  $$y=\frac{13}{11}$$
- step10: Substitute the value of $$y:$$
  $$x=\frac{5+5\times \frac{13}{11}}{4}$$
- step11: Calculate:
  $$x=\frac{30}{11}$$
- step12: Calculate:
  $$\left\{ \begin{array}{l}x=\frac{30}{11}\y=\frac{13}{11}\end{array}\right.$$
- step13: Check the solution:
  $$\left\{ \begin{array}{l}x=\frac{30}{11}\y=\frac{13}{11}\end{array}\right.$$
- step14: Rewrite:
  $$\left(x,y\right) = \left(\frac{30}{11},\frac{13}{11}\right)$$
Solve the system of equations $$ 4 x - 5 y = 5; 3 x + 10 y = -20 $$.
Solve the system of equations by following steps:
- step0: Solve using the substitution method:
  $$\left\{ \begin{array}{l}4x-5y=5\3x+10y=-20\end{array}\right.$$
- step1: Solve the equation:
  $$\left\{ \begin{array}{l}x=\frac{5+5y}{4}\3x+10y=-20\end{array}\right.$$
- step2: Substitute the value of $$x:$$
  $$3\times \frac{5+5y}{4}+10y=-20$$
- step3: Simplify:
  $$\frac{3\left(5+5y\right)}{4}+10y=-20$$
- step4: Multiply both sides of the equation by LCD:
  $$\left(\frac{3\left(5+5y\right)}{4}+10y\right)\times 4=-20\times 4$$
- step5: Simplify the equation:
  $$15+55y=-80$$
- step6: Move the constant to the right side:
  $$55y=-80-15$$
- step7: Subtract the numbers:
  $$55y=-95$$
- step8: Divide both sides:
  $$\frac{55y}{55}=\frac{-95}{55}$$
- step9: Divide the numbers:
  $$y=-\frac{19}{11}$$
- step10: Substitute the value of $$y:$$
  $$x=\frac{5+5\left(-\frac{19}{11}\right)}{4}$$
- step11: Calculate:
  $$x=-\frac{10}{11}$$
- step12: Calculate:
  $$\left\{ \begin{array}{l}x=-\frac{10}{11}\y=-\frac{19}{11}\end{array}\right.$$
- step13: Check the solution:
  $$\left\{ \begin{array}{l}x=-\frac{10}{11}\y=-\frac{19}{11}\end{array}\right.$$
- step14: Rewrite:
  $$\left(x,y\right) = \left(-\frac{10}{11},-\frac{19}{11}\right)$$
Let's analyze the systems of equations one by one to find which one has a solution approximately equal to $$ (-2.7, -1.2) $$.

1. **First System:**
   $$
   4x - 5y = -5 \
   3x + 10y = -20
   $$
   The solution is approximately $$ (-2.7, -1.2) $$.

2. **Second System:**
   $$
   4x - 5y = 5 \
   3x + 10y = 20
   $$
   The solution is approximately $$ (2.7, 1.2) $$.

3. **Third System:**
   $$
   4x - 5y = 5 \
   3x + 10y = -20
   $$
   The solution is approximately $$ (-0.9, -1.7) $$.

4. **Fourth System:**
   $$
   4x - 5y = -5 \
   3x + 10y = 20
   $$
   The solution is approximately $$ (0.9, 1.7) $$.

From the analysis, the first system of equations:
$$
4x - 5y = -5 \quad \text{and} \quad 3x + 10y = -20
$$
has a solution of approximately $$ (-2.7, -1.2) $$.

Thus, the correct answer is the first system.
The first system of equations $$ 4x - 5y = -5 $$ and $$ 3x + 10y = -20 $$ has a solution of approximately $$ (-2.7, -1.2) $$.

Which system of equations has a solution of approximately \( (-2.7,-1.2) \) ? \( 4 x-5 y=-5 \) and \( 3 x+10 y=-20 \) \( 4 x-5 y=5 \) and \( 3 x+10 y=20 \) \( 4 x-5 y=5 \) and \( 3 x+10 y=-20 \) \( 4 x-5 y=-5 \) and \( 3 x+10 y=20 \) Mark this and return

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