The given linear system is:
[tex]\displaystyle \left \{ {{5x+7y=33} \atop {11x+7y=39}} \right.[/tex]
The question is asking to solve using elimination. When eliminating, you can either eliminate x or y. In this system, y is much easier to eliminate. The y variable in both equations are 7y. To eliminate, one of them has to be negative, so multiply one of the equations by -1.
I will be multiplying the second equation by -1.
[tex]\displaystyle (11x + 7y = 39) \times -1 = -11x-7y=-39[/tex]
Rewrite the system:
[tex]\displaystyle \left \{ {{5x+7y=33} \atop {-11x-7y=-39}} \right.[/tex]
Subtract:
[tex]5x+7y=33\\-11x-7y=-39[/tex]
[tex]5x-11x=33-39\\-6x=-6[/tex]
Lastly, you need to leave the variable x alone. The variable is currently -6x or -6 times x. To remove it, you need to do the opposite of it, which is dividing by -6.
[tex]\displaystyle \frac{-6x}{-6} =\frac{-6}{-6}[/tex]
[tex]\displaystyle x=1[/tex]
Now that you have the value of x, substitute it into one of the equations to find y. I will be substituting it into the first equation.
[tex]5x+7y=33 \rightarrow 5(1)+7y=33[/tex]
Open the parentheses and multiply:
[tex]5+7y=33[/tex]
Move 5 to the other side to leave the variable alone:
[tex]5+7y-5=33-5[/tex]
You will be subtracting since you're "removing" it by doing the opposite of it.
[tex]7y=28[/tex]
Lastly, divide both sides by 7 to leave y alone.
[tex]\displaystyle \frac{7y}{7} =\frac{28}{7}[/tex]
[tex]y=4[/tex]
[tex]\displaystyle (x,y) \rightarrow (1, 4)[/tex]
The answer is (1, 4).
Answer:
(1, 4)
Tbh this question was a bit of a challenge for me but hopefully it helps ya.
Step-by-step explanation:
First, let's solve for y. So we have to eliminate the x values.
(5x + 7y = 33 ) *11
(11x + 7y = 39) *-5
___________
55x + 77y = 363
-55x -35y = -195 This eliminates the x value.
42y = 168 Divide both sides by 42 to isolate y: 168/42= y = 4
Now let's solve for x.
(11x + 7y = 39) *7
(5x + 7y = 33 ) *-7
___________
77x + 49y = 273
-35x -49y = -231 This cancels out the y values, which leaves 42x = 42
x = 1 and y = 4, so the answer is (1, 4)
