Respuesta :
Solving for the equation of the first line
Given: slope of 2, and passes through point (1,-1)
Recall the point slope form of a line
[tex]\begin{gathered} y-y_1=m(x-x_1) \\ \text{where} \\ m\text{ is the slope} \\ (x_1,y_1)\text{ is the point in the line} \end{gathered}[/tex]Substituting the given, and solving in terms of y, the equation of the first line is
[tex]\begin{gathered} y-(-1)=2\lbrack x-(1)\rbrack \\ y+1=2(x-1) \\ y+1=2x-2 \\ y=2x-2-1 \\ y=2x-3 \end{gathered}[/tex]Solving for the equation of the second line
Given: points (8,-5) and (10,-10)
First solve for the slope of the line
[tex]\begin{gathered} m=\frac{y_2-y_1}{x_2-x_1} \\ m=\dfrac{-10-(-5)}{10-8} \\ m=\dfrac{-10+5}{2} \\ m=-\frac{5}{2} \end{gathered}[/tex]Now that we have the slope, use the point-slope form and solve in terms of y to get the equation of the second line
[tex]\begin{gathered} \text{Using the point (10,-10) we have} \\ y-(-10)=-\frac{5}{2}(x-10) \\ y+10=-\frac{5}{2}x+25 \\ y=-\frac{5}{2}x+25-10 \\ y=-\frac{5}{2}x+15 \end{gathered}[/tex]Solving for the solution of the system
Now that we have two equations
[tex]\begin{cases}y=2x-3 \\ y=-\frac{5}{2}x+15\end{cases}[/tex]Equate both y's, substitute, and solve for x
[tex]\begin{gathered} y=y \\ 2x-3=-\frac{5}{2}x+15 \\ \\ \text{Multiply both sides by }2,\text{ to get rid of fractions in the right side of equation} \\ \Big(2x-3\Big)\cdot2=\Big(-\frac{5}{2}x+15\Big)\cdot2 \\ 4x-6=-5x+30 \\ \\ \text{Add both sides by }5x+6 \\ 4x-6=-5x+30 \\ 4x+5x-6+6=-5x+5x+30+6 \\ 9x\cancel{-6+6}=\cancel{-5x+5x}+36 \\ 9x=36 \\ \\ \text{Divide both sides by }9 \\ 9x=36 \\ \frac{9x}{9}=\frac{36}{9} \\ x=4 \end{gathered}[/tex]Now that we have solved for x, substitute the value to get the solution for y
Use either of the two equation, in this instance we will be using the first equation (using the second equation will work just as well)
[tex]\begin{gathered} y=2x-3 \\ y=2(4)-3 \\ y=8-3 \\ y=5 \end{gathered}[/tex]With x = 4, and y = 5, the solution to the system therefore is (4,5).
