Problem: Solve the following system of linear equations by substitution.{8x−3y=66x+12y=−24{8𝑥−3𝑦=66𝑥+12𝑦=−24Solution:This one looks a little more complicated than the one we just tried. Remember, with substitution we want to have one variable isolated (by itself), so if there is not one with a coefficient of 1 we'll need to do a little more work. There is no variable that has a coefficient of +1 or of –1 in this system. However, the second equation has coefficients and a constant that are multiples of 6. The second equation will be solved for the variable ‘x𝑥’.6x+12y=−246𝑥+12𝑦=−246x+12y−12y=−24−12y6𝑥+12𝑦−12𝑦=−24−12𝑦6x=−24−12y6𝑥=−24−12𝑦6x6=−246−12y66𝑥6=−246−12𝑦6x=−4−2y𝑥=−4−2𝑦Substitute (−4−2y)(−4−2𝑦) into the first equation for ‘x𝑥’.8x−3y=68𝑥−3𝑦=68(−4−2y)−3y=68(−4−2𝑦)−3𝑦=6Apply the distributive property and solve the equation.−32−16y−3y=6−32−16𝑦−3𝑦=6−32−19y=6−32−19𝑦=6−32+32−19y=6+32−32+32−19𝑦=6+32−19y=38−19𝑦=38−19y−19=38−19−19𝑦−19=38−19y=−2𝑦=−2Substitute –2 for y𝑦 into one of the equations. We can use x=−4−2y𝑥=−4−2𝑦 since it is one of the original equations in a different form.x=−4−2y𝑥=−4−2𝑦x=−4−2(𝑥=−4−2( Answer 1 Question 5 ))x=−4+𝑥=−4+ Answer 2 Question 5x=0𝑥=0The solution is (( Answer 3 Question 5 ,, Answer 4 Question 5 )).Check your answer to be sure no errors were made in the calculations.CheckQuestion 5