Exercise 101 Page 290

In this system of equations, at least one of the variables has a coefficient of 1. Therefore, we will approach its solution with the Substitution Method. When solving a system of equations using substitution, there are three steps.

1. Isolate a variable in one of the equations.
2. Substitute the expression for that variable into the other equation and solve.
3. Substitute the solution into one of the equations and solve for the value of the other variable.

Observing the given equations, it looks like it will be simplest to isolate u in the second equation.

5u+6v=2 & (I) 2v=u-10 & (II)

AddEqn

(II): LHS+10=RHS+10

5u+6v=2 2v+10=u

RearrangeEqn

(II): Rearrange equation

5u+6v=2 u=2v+10

Now that we've isolated u, we can solve the system by substitution.

5u+6v=2 u=2v+10

Substitute

(I):u= 2v+10

5( 2v+10)+6v=2 u=2v+10

▼

(I): Simplify

Distr

(I): Distribute 5

10v+50+6v=2 u=2v+10

AddTerms

(I): Add terms

16v+50=2 u=2v+10

SubEqn

(I):LHS-50=RHS-50

16v=- 48 u=2v+10

DivEqn

(I): .LHS /16.=.RHS /16.

v=- 3 u=2v+10

Great! Now, to find the value of u we need to substitute v=- 3 into either one of the equations in the given system. Let's use the second equation.

v=- 3 u=2v+10

Substitute

(II): v= - 3

v=- 3 u=2( - 3)+10

▼

(II): Simplify

MultPosNeg

(II): a(- b)=- a * b

v=- 3 u=- 6+10

AddTerms

(II): Add terms

v=- 3 u=4

The solution, or point of intersection, to this system of equations is the point (4, - 3).

Checking Our Answer

To check our answer, we will substitute our solution into both equations. If doing so results in true statements, then our solution is correct.

5u+6v=2 & (I) 2v=u-10 & (II)

(I), (II): u= 4, v= - 3

5( 4)+6( - 3)? =2 2( - 3)? = 4-10

Multiply

(I): Multiply

20+6(- 3)=2 2(- 3)=4-10

(I), (II): a(- b)=- a * b

20-18? =2 - 6? =4-10

(I), (II): Subtract terms

2=2 - 6 = - 6

Because both equations are true statements, we know that our solution is correct.