Review the following concepts if needed:

- Linear Systems in Two Variables for Intermediate Algebra
- Linear Systems in Three Variables for Intermediate Algebra
- Determinants and Cramer's Rule for Intermediate Algebra
- Matrix Methods for Intermediate Algebra
- Solving Nonlinear Systems for Intermediate Algebra

**7.1** Use the addition method to solve the following systems.

The coefficients of y in the equations are additive inverses. Therefore y will be eliminated if the equations are added.

Now substitute 4 for *x* in either of the original equations and solve for *y*. We choose *x* + y = –3, since *y* can be readily isolated in the equation.

Check: We substitute *x* = 4 and *y* = –7 into both equations.

The solution to the system is (4, –7). The solution set is {(4, –7)}. The equations are consistent and independent.

If the first equation is multiplied by 5, the coefficients of *t* are additive inverses. The variable *t* is eliminated when the equations are added.

Now substitute 2 for *s* in either of the original equations and solve for *t*. We choose 3*s* + *t* = 7, since *t* can be easily isolated.

Check: We substitute *s* = 2 and *t* = 1 into both equations.

The solution to the system is (2, 1). The solution set is {(2, 1)}. The equations are consistent and independent.

The second equation states the value of *y* directly. We need only to substitute *y* = 6 into the first equation and solve for *x*.

Check: Substitute into both equations.

The solution is (5, 6). The solution set is {(5, 6)}. The equations are consistent and independent.

Multiply the second equation by (–3) and add.

Now substitute *w* = 2 into either of the original equations and solve for *v*. We choose *v* + 2*w* = 7.

Check: Substitute into both equations.

The solution is (3, 2). The solution set is {(3, 2)}. The equations are consistent and independent.

We choose to eliminate *u* since the coefficients are opposite in sign. We therefore need not multiply by negative factors. Multiply the first equation by 2 and the second equation by 3 to obtain additive inverse coefficients of *u* in the equations.

Now substitute *t* = 2 into either of the original equations and solve for *u*. We choose 5*t* + 2*u* = 4 since the coefficient of *u* is positive.

Check: Substitute into both equations.

The solution is (2, –3). The solution set is {(2, –3)}. The equations are consistent and independent.

Our approach will be more apparent if we first clear fractions. Multiply the first equation by 6 and the second by 8.

The coefficients of *y* will be additive inverses if we now multiply 3*x* – 4*y* = 2 by 2.

Check: Substitute into the original equations.

Multiply the first equation by (–3).

There is no solution to the system. The solution set is { } = Ø. The lines are parallel. The system is inconsistent.

Multiply the first equation by (–3).

All of the ordered pairs which satisfy the first equation also satisfy the second. The equations represent the same line. The solution set consists of the infinitely many ordered pairs which satisfy either equation. The equations are dependent.

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